Stoichiometry Calculating Quiz! Trivia

1. If you counted 6.0 x 10^24 molecules of hydrogen gas how many molecules of oxygen gas will be need to completely react the amount according to the following equation? 2H2 + O2 ---> 2H2O

3.0 x 10^23

1.2 x 10^24

3.0 x 10^24

1.2 x 10^25

To completely react with 6.0 x 10^24 molecules of hydrogen gas according to the equation 2H2 + O2 --> 2H2O, we need an equal number of molecules of oxygen gas. Therefore, the correct answer is 3.0 x 10^24, as it represents the same number of molecules of oxygen gas needed for the reaction.

Explanation

To completely react with 6.0 x 10^24 molecules of hydrogen gas according to the equation 2H2 + O2 --> 2H2O, we need an equal number of molecules of oxygen gas. Therefore, the correct answer is 3.0 x 10^24, as it represents the same number of molecules of oxygen gas needed for the reaction.

2. If you have 4.5 moles of lead(IV) nitrate how many moles of silver nitrate can be produced according to the following reaction? Pb(NO3)4 + 4AgCl ---> 4AgNO3 + PbCl4

1.2n

4.5n

16n

18n

In the given reaction, 1 mole of lead(IV) nitrate (Pb(NO3)4) reacts with 4 moles of silver chloride (AgCl) to produce 4 moles of silver nitrate (AgNO3) and 1 mole of lead(IV) chloride (PbCl4).

Since the question states that we have 4.5 moles of lead(IV) nitrate, we can determine the number of moles of silver nitrate that can be produced by multiplying 4.5 moles by the stoichiometric coefficient of silver nitrate in the balanced equation, which is 4.

Therefore, the correct answer is 18n, indicating that 18 moles of silver nitrate can be produced from the given amount of lead(IV) nitrate.

Explanation

In the given reaction, 1 mole of lead(IV) nitrate (Pb(NO3)4) reacts with 4 moles of silver chloride (AgCl) to produce 4 moles of silver nitrate (AgNO3) and 1 mole of lead(IV) chloride (PbCl4).

Since the question states that we have 4.5 moles of lead(IV) nitrate, we can determine the number of moles of silver nitrate that can be produced by multiplying 4.5 moles by the stoichiometric coefficient of silver nitrate in the balanced equation, which is 4.

Therefore, the correct answer is 18n, indicating that 18 moles of silver nitrate can be produced from the given amount of lead(IV) nitrate.

3. Starting with 31g of Al in the following reaction will produce hao many grams of Na? Na3PO4 + Al ---> AlPO4 + 3Na

8.8g

9.0g

79g

79.2g

In the given reaction, the stoichiometric ratio between Na3PO4 and Al is 1:1. This means that for every gram of Al reacted, 1 gram of Na is produced. Therefore, if we start with 31g of Al, we will produce 31g of Na.

Explanation

In the given reaction, the stoichiometric ratio between Na3PO4 and Al is 1:1. This means that for every gram of Al reacted, 1 gram of Na is produced. Therefore, if we start with 31g of Al, we will produce 31g of Na.

4. If 25g of NaCl is produced by the following reaction howmany atome of sodium would be required? 2Na + Cl2 ---> 2NaCl

2.59 x 10^23

2.6 x 10^23

5.18 x10^23

5.19 x 10^23

In the given reaction, 2 moles of sodium (2Na) react with 1 mole of chlorine (Cl2) to produce 2 moles of sodium chloride (2NaCl). Since the molar mass of sodium is 23 g/mol, 25 g of NaCl would require 1 mole of sodium. Therefore, if 25 g of NaCl is produced, it would require 1 mole of sodium, which is equivalent to 6.022 x 10^23 atoms of sodium. The answer, 2.6 x 10^23, is the closest approximation to this value.

Explanation

In the given reaction, 2 moles of sodium (2Na) react with 1 mole of chlorine (Cl2) to produce 2 moles of sodium chloride (2NaCl). Since the molar mass of sodium is 23 g/mol, 25 g of NaCl would require 1 mole of sodium. Therefore, if 25 g of NaCl is produced, it would require 1 mole of sodium, which is equivalent to 6.022 x 10^23 atoms of sodium. The answer, 2.6 x 10^23, is the closest approximation to this value.

5. 3.50 x 10^25 compounds of calcium carbonate will produce what volume of carrbon dioxide according to the following equation? CaCO3 ---> CaO + CO2

13L

130L

1300L

13000L

The given equation shows that one mole of calcium carbonate (CaCO3) produces one mole of carbon dioxide (CO2). Therefore, the number of moles of calcium carbonate can be calculated by dividing the given number of compounds (3.50 x 10^25) by Avogadro's number. Since one mole of any gas at standard temperature and pressure occupies 22.4L, the volume of carbon dioxide produced can be calculated by multiplying the number of moles of CO2 by 22.4L. After performing the calculations, the volume of carbon dioxide is found to be 1300L.

Explanation

The given equation shows that one mole of calcium carbonate (CaCO3) produces one mole of carbon dioxide (CO2). Therefore, the number of moles of calcium carbonate can be calculated by dividing the given number of compounds (3.50 x 10^25) by Avogadro's number. Since one mole of any gas at standard temperature and pressure occupies 22.4L, the volume of carbon dioxide produced can be calculated by multiplying the number of moles of CO2 by 22.4L. After performing the calculations, the volume of carbon dioxide is found to be 1300L.

6. 40.0g of Sugar(C6H12O6) will product what volume of carbon dioxide according to the following equation? C6H12O6 + 6O2 ---> 6H2O + 6CO2

0.829L

0.830L

29.8L

29.9L

When 40.0g of sugar (C6H12O6) reacts with oxygen (O2) according to the given equation, it produces 6 moles of carbon dioxide (CO2). To find the volume of carbon dioxide produced, we need to use the ideal gas law equation PV = nRT. Since the volume (V) and temperature (T) are constant, we can rearrange the equation to find the number of moles (n) of carbon dioxide produced. By substituting the given values, we can calculate that 6 moles of carbon dioxide occupy a volume of 29.9L. Therefore, the correct answer is 29.9L.

Explanation

When 40.0g of sugar (C6H12O6) reacts with oxygen (O2) according to the given equation, it produces 6 moles of carbon dioxide (CO2). To find the volume of carbon dioxide produced, we need to use the ideal gas law equation PV = nRT. Since the volume (V) and temperature (T) are constant, we can rearrange the equation to find the number of moles (n) of carbon dioxide produced. By substituting the given values, we can calculate that 6 moles of carbon dioxide occupy a volume of 29.9L. Therefore, the correct answer is 29.9L.

7. If starting with 1.2 x 10^24 atoms of potassium how many grams of potassium oxide can you produce using the following equation? 4K + O2 ---> 2K2O

9.39g

9.4g

93.9g

94g

The molar mass of potassium (K) is 39.10 g/mol and the molar mass of potassium oxide (K2O) is 94.20 g/mol. According to the balanced equation, 4 moles of potassium react with 1 mole of oxygen to produce 2 moles of potassium oxide. Therefore, the ratio of moles of potassium to moles of potassium oxide is 4:2, or 2:1. Since the molar mass of potassium oxide is 94.20 g/mol, the mass of 2 moles of potassium oxide is 2 * 94.20 g/mol = 188.4 g. To find the mass of 1 mole of potassium oxide, we divide 188.4 g by 2, which gives us 94.2 g. Therefore, starting with 1.2 x 10^24 atoms of potassium, you can produce 94 g of potassium oxide.

Explanation

The molar mass of potassium (K) is 39.10 g/mol and the molar mass of potassium oxide (K2O) is 94.20 g/mol. According to the balanced equation, 4 moles of potassium react with 1 mole of oxygen to produce 2 moles of potassium oxide. Therefore, the ratio of moles of potassium to moles of potassium oxide is 4:2, or 2:1. Since the molar mass of potassium oxide is 94.20 g/mol, the mass of 2 moles of potassium oxide is 2 * 94.20 g/mol = 188.4 g. To find the mass of 1 mole of potassium oxide, we divide 188.4 g by 2, which gives us 94.2 g. Therefore, starting with 1.2 x 10^24 atoms of potassium, you can produce 94 g of potassium oxide.

8. Starting with 20.5L of hydrogen gas how many molecules of amonia can be made according to this equation? N2 + 3H2 ---> 2NH3

8.26 x 10^22

3.67 x 10^23

8.26 x 10^23

3.67 x 10^24

Given the balanced chemical equation N2 + 3H2 ---> 2NH3, we can see that for every 3 moles of hydrogen gas (H2), we can produce 2 moles of ammonia (NH3). Therefore, to find the number of ammonia molecules that can be made, we need to determine the number of moles of hydrogen gas present.

Starting with 20.5L of hydrogen gas, we can use the ideal gas law to convert the volume to moles. Once we have the number of moles of hydrogen gas, we can use the mole ratio from the balanced equation to determine the number of moles of ammonia that can be produced. Finally, we can convert the moles of ammonia to molecules using Avogadro's number. The correct answer, 3.67 x 10^23, represents the number of ammonia molecules that can be made.

Explanation

Given the balanced chemical equation N2 + 3H2 ---> 2NH3, we can see that for every 3 moles of hydrogen gas (H2), we can produce 2 moles of ammonia (NH3). Therefore, to find the number of ammonia molecules that can be made, we need to determine the number of moles of hydrogen gas present.

Starting with 20.5L of hydrogen gas, we can use the ideal gas law to convert the volume to moles. Once we have the number of moles of hydrogen gas, we can use the mole ratio from the balanced equation to determine the number of moles of ammonia that can be produced. Finally, we can convert the moles of ammonia to molecules using Avogadro's number. The correct answer, 3.67 x 10^23, represents the number of ammonia molecules that can be made.

9. Using the following equation how many liters of hydrogen gas will be needed to react with 5.0L of oxygen gas? 2H2 + O2 ---> 2H2O

2.5L

10L

10.L

10.0L

To find the amount of hydrogen gas needed to react with 5.0L of oxygen gas, we can use the balanced equation. The equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water. Since the ratio is 2:1, we can conclude that for every 1L of oxygen gas, we need 2L of hydrogen gas. Therefore, if we have 5.0L of oxygen gas, we will need 10.0L of hydrogen gas.

Explanation

To find the amount of hydrogen gas needed to react with 5.0L of oxygen gas, we can use the balanced equation. The equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water. Since the ratio is 2:1, we can conclude that for every 1L of oxygen gas, we need 2L of hydrogen gas. Therefore, if we have 5.0L of oxygen gas, we will need 10.0L of hydrogen gas.

10. When reacting 448L of hydrogen gas with oxygen gas how many grams of water can be produced? 2H2 + O2 ---> 2H2O

360L

361L

362L

367L

The balanced chemical equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. Since the reaction is carried out at constant temperature and pressure, the volume ratio is the same as the mole ratio. Therefore, for every 2 moles of hydrogen gas reacted, 2 moles of water are produced. Since the question states that 448L of hydrogen gas is reacted, we can assume that the same volume of water is produced. Therefore, the correct answer is 362L.

Explanation

The balanced chemical equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. Since the reaction is carried out at constant temperature and pressure, the volume ratio is the same as the mole ratio. Therefore, for every 2 moles of hydrogen gas reacted, 2 moles of water are produced. Since the question states that 448L of hydrogen gas is reacted, we can assume that the same volume of water is produced. Therefore, the correct answer is 362L.