An Easy Quiz On Stoichiometry

Explanation
In the balanced equation, the stoichiometric ratio between sodium (Na) and chlorine gas (Cl2) is 2:1. This means that for every 2 moles of sodium, 1 mole of chlorine gas is required for complete reaction. Since we have 5.0 moles of sodium, we can calculate the moles of chlorine gas needed by dividing 5.0 by 2. This gives us 2.5 moles of chlorine gas as the answer.

Explanation
When C8H18 is combusted, it reacts with O2 to produce CO2 and H2O. The balanced equation for this reaction is C8H18 + 12.5O2 -> 8CO2 + 9H2O. From the balanced equation, we can see that for every 1 mole of C8H18 combusted, 9 moles of H2O are produced. To find the mass of water produced, we need to convert the given mass of C8H18 to moles using its molar mass, then use the mole ratio from the balanced equation to find the moles of water produced, and finally convert the moles of water to grams using the molar mass of water. The correct answer is 37.8 grams.

Explanation
In the given balanced chemical equation, the stoichiometric ratio between CH4 and O2 is 1:1. This means that for every 1 mole of CH4, 1 mole of O2 is required to completely burn it. Since we are given 0.5 moles of CH4, we can conclude that 0.5 moles of O2 is required to burn it. Therefore, the answer is 1 mole of O2.

Explanation
In the balanced chemical equation, the coefficient of NaOH is 1, which means that 1 mole of NaOH reacts with 1 mole of HCl. To find the number of grams of NaOH that will react with 36.5 g of HCl, we need to convert the grams of HCl to moles using its molar mass (36.5 g/mol). Since the molar mass of NaOH is 40 g/mol, we can conclude that 36.5 g of HCl will react with 40 g of NaOH. Therefore, the correct answer is 40 grams.

Explanation
The balanced equation for the combustion of hydrogen gas in oxygen gas is 2H2 + O2 → 2H2O. This means that for every 2 moles of hydrogen gas, 2 moles of water are formed. Since there are 5 moles of hydrogen gas, we can calculate that 5 moles of water are formed. Therefore, the answer is 5.

Explanation
The balanced reaction for the combustion of methane gas in oxygen is CH4 + 2O2 --> CO2 + 2H2O. This equation shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water vapor. This balanced equation follows the law of conservation of mass, as the number of atoms of each element is the same on both sides of the equation.

Explanation
When methane gas (CH4) is burned in oxygen (O2), it reacts to form carbon dioxide (CO2) and water vapor (H2O). The balanced chemical equation for this reaction is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that for every 1 mole of methane burned, 1 mole of carbon dioxide is produced. The molar mass of methane is 16.04 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol.

To find the mass of carbon dioxide produced when 10 grams of methane is burned, we can use the molar mass ratio:

(10 g CH4) x (1 mol CH4 / 16.04 g CH4) x (1 mol CO2 / 1 mol CH4) x (44.01 g CO2 / 1 mol CO2) = 27.4 g CO2

Therefore, 27.4 grams of CO2 will be produced when 10 grams of methane is burned.

Explanation
To produce 10.0 moles of oxygen gas in the decomposition of water, 20.0 moles of water are needed. This is because the balanced chemical equation for the decomposition of water is 2H2O -> 2H2 + O2. From the equation, we can see that for every 2 moles of water, 1 mole of oxygen gas is produced. Therefore, to produce 10.0 moles of oxygen gas, we need twice that amount of water, which is 20.0 moles.

Explanation
To determine the number of moles of water needed to produce 32 grams of oxygen gas, we need to use the molar mass of oxygen. The molar mass of oxygen is 16 grams per mole. Since the molar ratio between water and oxygen gas is 2:1, we can calculate the number of moles of water by dividing the mass of oxygen gas by the molar mass of oxygen. Therefore, 32 grams of oxygen gas is equal to 2 moles of oxygen gas, which means that 2 moles of water are needed to produce 32 grams of oxygen gas.