76 report sheet · chemical reactions of copper and percent yield\nquestions\n1. when zinc (or aluminum) was…

76 report sheet · chemical reactions of copper and percent yield\nquestions\n1. when zinc (or aluminum) was allowed to react with the copper sulfate, what was the limiting reagent?\n2. if your percent yield of copper was greater than 100%, what are three plausible errors you may have made?\n3. consider the combustion of methane, ch₄:\nch₄(g)+2o₂(g)→co₂(g)+2h₂o(g)\nsuppose 2.8 mol of methane are allowed to react with 5.0 mol of oxygen.\n(a) what is the limiting reagent?\n(b) how many moles of co₂ can be made from this reaction? how many grams of co₂?\n4. 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?

76 report sheet · chemical reactions of copper and percent yield\nquestions\n1. when zinc (or aluminum) was allowed to react with the copper sulfate, what was the limiting reagent?\n2. if your percent yield of copper was greater than 100%, what are three plausible errors you may have made?\n3. consider the combustion of methane, ch₄:\nch₄(g)+2o₂(g)→co₂(g)+2h₂o(g)\nsuppose 2.8 mol of methane are allowed to react with 5.0 mol of oxygen.\n(a) what is the limiting reagent?\n(b) how many moles of co₂ can be made from this reaction? how many grams of co₂?\n4. 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?

Answer

Explanation:

Question 1

Step1: Write the reaction equations

For zinc: $Zn + CuSO_4\rightarrow ZnSO_4+Cu$; for aluminum: $2Al + 3CuSO_4\rightarrow Al_2(SO_4)_3 + 3Cu$. To determine the limiting - reagent, we need to know the amounts of reactants. If we assume equal molar amounts of zinc/aluminum and copper sulfate are given initially, for zinc, 1 mole of $Zn$ reacts with 1 mole of $CuSO_4$, and for aluminum, 2 moles of $Al$ react with 3 moles of $CuSO_4$. Without specific amounts, if we consider standard conditions where we compare molar ratios, if the amount of $Zn$ or $Al$ is less than the stoichiometrically - required amount based on the amount of $CuSO_4$, then $Zn$ or $Al$ is the limiting reagent. If the amount of $CuSO_4$ is less, then $CuSO_4$ is the limiting reagent.

Question 2

Step1: Contamination

Contamination of the product with an impurity that increases the mass measured can cause a percent yield greater than 100%. For example, if there are un - reacted reactants or other substances mixed in with the copper product during weighing.

Step2: Incomplete drying

If the copper product is not completely dried before weighing, the extra mass of water can make the measured mass of copper higher than the actual mass of pure copper formed, leading to a percent yield greater than 100%.

Step3: Measurement error

Errors in measuring the mass of the reactants or the product can also cause this. For example, an incorrect zero - setting on the balance or misreading the scale can result in inaccurate mass measurements.

Question 3 (a)

Step1: Determine the mole ratio

The balanced equation is $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$. The mole ratio of $CH_4$ to $O_2$ is 1:2. Given $n(CH_4) = 2.8$ mol and $n(O_2)=5.0$ mol.

Step2: Calculate the amount of $O_2$ required for complete reaction of $CH_4$

Based on the mole ratio, for 2.8 mol of $CH_4$, the amount of $O_2$ required is $n_{O_2\ required}=2\times n(CH_4)=2\times2.8 = 5.6$ mol. Since we have only 5.0 mol of $O_2$, $O_2$ is the limiting reagent.

Question 3 (b)

Step1: Calculate moles of $CO_2$

Since $O_2$ is the limiting reagent, from the balanced equation, the mole ratio of $O_2$ to $CO_2$ is 2:1. Given $n(O_2) = 5.0$ mol, the moles of $CO_2$ produced, $n(CO_2)=\frac{1}{2}\times n(O_2)=\frac{1}{2}\times5.0 = 2.5$ mol.

Step2: Calculate grams of $CO_2$

The molar mass of $CO_2$ is $M(CO_2)=44.01$ g/mol. So the mass of $CO_2$, $m(CO_2)=n(CO_2)\times M(CO_2)=2.5\ mol\times44.01\ g/mol = 110.025$ g.

Question 4

Step1: Calculate moles of reactants

The molar mass of $CH_4$ is $M(CH_4)=16.04$ g/mol, so $n(CH_4)=\frac{8.00\ g}{16.04\ g/mol}\approx0.499$ mol. The molar mass of $O_2$ is $M(O_2) = 32.00$ g/mol, so $n(O_2)=\frac{16.00\ g}{32.00\ g/mol}=0.5$ mol.

Step2: Determine the limiting reagent

The mole ratio of $CH_4$ to $O_2$ is 1:2. For 0.499 mol of $CH_4$, the amount of $O_2$ required is $2\times0.499 = 0.998$ mol. Since we have only 0.5 mol of $O_2$, $O_2$ is the limiting reagent.

Step3: Calculate moles of products and remaining reactants

From the balanced equation, the mole ratio of $O_2$ to $CO_2$ is 2:1 and $O_2$ to $H_2O$ is 1:1. Moles of $CO_2$ produced, $n(CO_2)=\frac{1}{2}\times n(O_2)=\frac{1}{2}\times0.5 = 0.25$ mol, and $m(CO_2)=n(CO_2)\times M(CO_2)=0.25\ mol\times44.01\ g/mol = 11.0025$ g. Moles of $H_2O$ produced, $n(H_2O)=n(CO_2) = 0.25$ mol, and $m(H_2O)=n(H_2O)\times M(H_2O)=0.25\ mol\times18.02\ g/mol = 4.505$ g. The amount of $CH_4$ remaining: $n_{CH_4\ remaining}=0.499-\frac{0.5}{2}=0.499 - 0.25=0.249$ mol, $m_{CH_4\ remaining}=0.249\ mol\times16.04\ g/mol\approx4.00$ g. The amount of $O_2$ remaining is 0 g since it is the limiting reagent.

Answer:

  1. Without specific amounts of reactants, if the amount of zinc/aluminum is less than the stoichiometrically - required amount based on the amount of $CuSO_4$, then $Zn$ or $Al$ is the limiting reagent. If the amount of $CuSO_4$ is less, then $CuSO_4$ is the limiting reagent.
  2. Contamination, incomplete drying, measurement error.
  3. (a) $O_2$ is the limiting reagent. (b) 2.5 mol of $CO_2$, 110.025 g of $CO_2$.
  4. $m(CH_4)\approx4.00$ g, $m(O_2) = 0$ g, $m(CO_2)=11.0025$ g, $m(H_2O)=4.505$ g.