Let's Practice Some Fun Stoichiometry Problems!

Explanation
The balanced chemical equation for the reaction between ammonia, copper (II) oxide, solid copper, water, and nitrogen gas is 2NH3 + 3CuO → 3Cu + 3H2O + N2. This equation shows that 2 moles of ammonia react with 3 moles of copper (II) oxide to produce 3 moles of solid copper, 3 moles of water, and 1 mole of nitrogen gas. Therefore, the coefficients for the balanced chemical equation are 2, 3, 3, 3, 1.

Explanation
In order to determine the limiting reactant, we need to compare the amount of reactants used with their respective molar masses. The molar mass of ammonia (NH3) is 17.03 g/mol, while the molar mass of copper (II) oxide (CuO) is 79.55 g/mol. By dividing the given mass of each reactant by their respective molar masses, we can calculate the number of moles present. The number of moles of ammonia is 2.35 mol, while the number of moles of copper (II) oxide is 1.01 mol. Since the ratio of ammonia to copper (II) oxide in the balanced equation is 4:1, we can see that there is an excess of copper (II) oxide and it is the limiting reactant.

Explanation
The mass of nitrogen gas produced in the reaction is determined to be 9.4 g.

Explanation
The excess reactant is ammonia because it is not fully consumed in the reaction. In a chemical reaction, the reactant that is present in a greater amount than necessary is called the excess reactant. In this case, ammonia is in excess, meaning there is more ammonia than required to react with the copper(II) oxide.

Explanation
Ammonia is the excess reactant in the reaction, which means it will not be completely used up. To determine how much is left over, a mass-mass problem can be used. The given amount of copper(II) oxide is 80g, and the unknown is the grams of ammonia. The answer provided, 28.6g, represents the amount of ammonia that is left over after the reaction.

Explanation
When copper(II) oxide is heated with hydrogen gas, a chemical reaction takes place resulting in the formation of water and copper metal. The balanced equation for this reaction is: CuO + H2 -> H2O + Cu.

To determine the mass of copper that can be obtained, we need to calculate the molar mass of copper(II) oxide (CuO), which is 63.55 g/mol for copper and 16.00 g/mol for oxygen. Adding these together gives us a molar mass of 79.55 g/mol for CuO.

Next, we need to calculate the number of moles of CuO present in 32.0 g of CuO. This can be done by dividing the given mass by the molar mass: 32.0 g / 79.55 g/mol = 0.402 mol.

According to the balanced equation, the molar ratio between CuO and Cu is 1:1. This means that for every 1 mole of CuO, we will obtain 1 mole of Cu.

Therefore, the mass of copper that can be obtained is equal to the molar mass of copper (63.55 g/mol) multiplied by the number of moles of CuO: 0.402 mol * 63.55 g/mol = 25.5 g.

Hence, the correct answer is 25.5g.

Explanation
The balanced equation for the reaction between nitrogen monoxide and oxygen gas to produce nitrogen dioxide is 2NO + O2 → 2NO2. This equation shows that 2 molecules of nitrogen monoxide react with 1 molecule of oxygen gas to produce 2 molecules of nitrogen dioxide. Therefore, the coefficients that balance the equation are 2, 1, 2.

Explanation
The molar mass of water (H2O) is 18.015 g/mol. From the balanced chemical equation for the decomposition of water (2H2O -> 2H2 + O2), we can see that for every 2 moles of water, we get 2 moles of hydrogen gas (H2). Therefore, we can calculate the number of moles of water in 36.0 g using the formula: moles = mass/molar mass. This gives us 36.0 g / 18.015 g/mol = 1.998 moles of water. Since the ratio of water to hydrogen gas is 2:2, the number of moles of hydrogen gas produced is also 1.998 moles. Finally, we can convert moles of hydrogen gas to grams using the molar mass of hydrogen gas (2.02 g/mol), giving us 1.998 moles * 2.02 g/mol = 4.04 g H2.

Explanation
The percent yield of hydrogen gas can be calculated by dividing the actual yield (3.80g) by the theoretical yield and multiplying by 100. Since no information is provided about the theoretical yield, we can assume that it is equal to the actual yield. Therefore, the percent yield would be 100%. However, since the given answer is 94%, it suggests that there may have been some loss or inefficiency in the collection process, resulting in a lower yield than expected.