76 Repot Sheet Chemical Reactioas of Copper and Percent Yield QUESTIONS 1. When

Question

76 Repot Sheet Chemical Reactioas of Copper and Percent Yield QUESTIONS 1. When zine (or aluminum) was allowed to react with the copper was the limiting reagent? e copper sulfase, what have 2. If your percent yield of copper was greater than 100%, what are three plausible errors you errors you may made? 3. Consider the combustion of methane, CH CH,(e) 20, ()cO,(g)+2H,O) 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 CO, can be made from this reaction? How many grams of Co,? 4. Suppose 8.00 g of CHs is allowed to burn in the presence of 16.00 g of oxygen. (See reaction in 3 above.) How much (in grams) CH4, 02. CO and HO (in grams) remain after the reaction is complete? Copyright © 2015 Pearson Education, Inc.

Explanation / Answer

1. Zn + CuSO4 = Cu + Zn(SO4)2

Limiting reagent would depend on the number of moles which are in the reaction.

For example in the reaction if zinc is still present after the reaction then CuSO4 is the limiting reagent if not then it is opposite.

2.  The mass of the copper measured before the experiment was too low.

The mass of the copper measured after the experiment was too high.

Contamination in this experiment with copper from another source--you didn't clean the equipment.

3. 1 mol of methane requires 2 mol of oxygen.

Divide the no of moles with their coefficients

CH4 = 2.8 mol

oxygen = 5/2 = 2.5

Since oxygen available is less than methane oxygen is the limiting reagent

Molecular mass MwCO2= 44 g/mol

mass=mol*Mw=2.5*44=110g of CO2

ratio 2 mol O2 is to 1 mol CO2, 2:1

Moles CO2=5*1/2=2.5 mol CO2

4. CH4 + O2 = CO2 + 2H2O

8 g of methane = 0.5 mol

16 g of oxygen = 0.5 mol

after reaction is complete 0.5 mol of CO2 = 22 g and 1 mol of water = 18 g.

5) H2SO4 + CuO = CuSO4 + H2

since one mole of sulfuric acid reacts with one mole of copper oxide

Molarity = no of moles / volume in L

4 x V(L) = 1.60/79

V = 0.00506 L = 5.06 mL

6) Zn + CuSO4 = ZnSO4 + Cu

moles of Zn : 2/65.0 = 0.0307

moles of CuSO4 = 2/159 = 0.0125

moles of zinc in excess = 0.0307 - 0.0125 = 0.0182

mass of zinc left = 0.0182 x 65 = 1.183 g

CuSO4 is completely consumed

Amount of ZnSO4 formed = 0.0125 mol = 0.0125 x 161 g = 2.0125 g

Amount of copper = 0.0125 x 63.5 = 0.793 g

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.