Question

76 report sheet · chemical reactions of copper and percent yield
questions

1. when zinc (or aluminum) was allowed to react with the copper sulfate, what was the limiting reagent?
2. if your percent yield of copper was greater than 100%, what are three plausible errors you may have made?
3. consider the combustion of methane, ch₄:

ch₄(g)+2o₂(g)→co₂(g)+2h₂o(g)
suppose 2.8 mol of methane are allowed to react with 5.0 mol of oxygen.
(a) what is the limiting reagent?
(b) how many moles of co₂ can be made from this reaction? how many grams of co₂?

1. suppose 8.00 g of ch₄ is allowed to burn in the presence of 16.00 g of oxygen. (see reaction in 3 above.) how much (in grams) ch₄, o₂, co₂, and h₂o (in grams) remain after the reaction is complete?

Explanation:

Step1: Determine limiting reagent for zinc/ aluminum - copper sulfate reaction

The reaction between zinc (or aluminum) and copper sulfate is a single - displacement reaction. To find the limiting reagent, we need to know the amounts (moles) of each reactant. Without specific amounts given in the problem statement for this part, we assume a general case. If we consider the balanced chemical equations:

• For zinc: $Zn + CuSO_4

ightarrow ZnSO_4+Cu$

• For aluminum: $2Al + 3CuSO_4

ightarrow Al_2(SO_4)_3 + 3Cu$
The limiting reagent is the reactant that is completely consumed first. If we have equal moles of zinc and copper sulfate, zinc is the limiting reagent in the first reaction as 1 mole of zinc reacts with 1 mole of copper sulfate. For aluminum, 2 moles of aluminum react with 3 moles of copper sulfate, so depending on the initial amounts, either aluminum or copper sulfate can be the limiting reagent.

Step2: Reasons for percent yield > 100% for copper

• Contamination: The copper product might be contaminated with an impurity that adds to its mass. For example, if there is unreacted zinc or aluminum powder mixed with the copper, it will increase the measured mass of the 'copper' product.
• Incomplete drying: If the copper is not completely dried before weighing, the presence of water (either adsorbed or as a hydrate) will increase the measured mass, leading to a calculated percent yield greater than 100%.
• Side - reactions: There could be side - reactions that produce additional copper - containing compounds that are weighed along with the copper, inflating the mass of the recovered product.

Step3: Determine limiting reagent for methane - oxygen reaction

The balanced chemical equation is $CH_4(g)+2O_2(g)
ightarrow CO_2(g)+2H_2O(g)$
We have 2.8 mol of $CH_4$ and 5.0 mol of $O_2$.
The mole ratio of $CH_4$ to $O_2$ is 1:2.
For 2.8 mol of $CH_4$, we would need $2.8\times2 = 5.6$ mol of $O_2$. But we only have 5.0 mol of $O_2$.
For 5.0 mol of $O_2$, we would need $\frac{5.0}{2}=2.5$ mol of $CH_4$.
Since we don't have enough $O_2$ to react with all of the $CH_4$, $O_2$ is the limiting reagent.

Step4: Calculate moles and mass of $CO_2$

From the balanced equation, the mole ratio of $O_2$ to $CO_2$ is 2:1.
Since we have 5.0 mol of $O_2$, the moles of $CO_2$ produced is $\frac{5.0}{2}=2.5$ mol.
The molar mass of $CO_2$ is $M = 12.01+2\times16.00=44.01$ g/mol.
The mass of $CO_2$ is $m = n\times M=2.5\ mol\times44.01\ g/mol = 110.025$ g

Step5: Calculate remaining amounts for $CH_4$, $O_2$, $CO_2$, $H_2O$ in fourth problem

First, convert masses to moles:

• Molar mass of $CH_4$ is $12.01 + 4\times1.01=16.05$ g/mol. So, 8.00 g of $CH_4$ is $\frac{8.00\ g}{16.05\ g/mol}=0.499$ mol
• Molar mass of $O_2$ is $2\times16.00 = 32.00$ g/mol. So, 16.00 g of $O_2$ is $\frac{16.00\ g}{32.00\ g/mol}=0.5$ mol

From the balanced equation $CH_4(g)+2O_2(g)
ightarrow CO_2(g)+2H_2O(g)$
The mole ratio of $CH_4$ to $O_2$ is 1:2.
For 0.5 mol of $O_2$, we need $\frac{0.5}{2}=0.25$ mol of $CH_4$.

• Remaining $CH_4$: $0.499 - 0.25=0.249$ mol, and its mass is $0.249\ mol\times16.05\ g/mol\approx4.00$ g
• $O_2$ is completely consumed, so 0 g of $O_2$ remains.
• Moles of $CO_2$ produced: Since the mole ratio of $O_2$ to $CO_2$ is 2:1, for 0.5 mol of $O_2$, we produce 0.25 mol of $CO_2$. The mass of $CO_2$ is $0.25\ mol\times44.01\ g/mol = 11.0025$ g
• Moles of $H_2O$ produced: The mole ratio of $O_2$ to $H_2O$ is 1:1, so 0.5 mol of $H_2O$ is produced. The molar mass of $H_2O$ is $2\times1.01+16.00 = 18.02$ g/mol. The mass of $H_2O$ is $0.5\ mol\ti…

Answer:

1. Depending on amounts, either zinc/aluminum or copper sulfate can be the limiting reagent. For equal moles of zinc and copper sulfate, zinc is limiting. For aluminum, it depends on initial mole - ratios.
2. Contamination, incomplete drying, side - reactions.
3. (a) $O_2$ is the limiting reagent. (b) 2.5 mol of $CO_2$, 110.025 g of $CO_2$.
4. $CH_4$: 4.00 g, $O_2$: 0 g, $CO_2$: 11.0025 g, $H_2O$: 9.01 g