In a laboratory experiment, 27.5 g of Cu were reacted with 125 g of HNO3 (63.01 g/mol):
3 Cu + 8 HNO 3 Cu(NO3)2 + 4H2O + 2 NO
a. How many grams of Cu(NO3)2 (187.57 g/mol) can be theoretically produced?
b. How many grams of excess reactant remain?
c. If the percent yield of Cu(NO3)2 is 87.3%, what is the actual yield in grams?

a) The theoretical yield of Cu(NO3)2 is 81.2 grams

b) There remains 52.2 grams HNO3

c) The actual yield is 70.9 grams Cu(NO3)2

Explanation:

Step 1: Data given

Mass of Cu = 27.5 grams

Mass of HNO3 = 125 grams

Atomic mass Cu = 63.546 g/mol

Molar mass of HNO3 = 63.01 g/mol

Step 2: The balanced equation

3 Cu + 8 HNO3 → 3Cu(NO3)2 + 4H2O + 2NO

Step 3: Calculate moles

Moles = mass / molar mass

Moles Cu = 27.5 grams / 63.546 g/mol

Moles Cu = 0.433 moles

Moles HNO3 = 125 grams / 63.01 g/mol

Moles HNO3 = 1.984 moles

Step 4: Calculate the limiting reactant

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2 , 4 moles H2O and 2 moles NO

Cu is the limiting reactant. It will completely be consumed. (0.433 moles). HNO3 is in excess. There will react 8/3 * 0.433 = 1.155 moles

There will remain 1.1984 - 1.155 = 0.829 moles

Step 5: Calculate mass HNO3 remaining

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.829 moles * 63.01 g/mol

Mass HNO3 = 52.2 grams

Step 6: Calculate moles Cu(NO3)2

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2 , 4 moles H2O and 2 moles NO

For 0.433 moles Cu, we'll have 0.433 moles Cu(NO3)2

Step 7: Calculate mass Cu(NO3)2

Mass Cu(NO3)2 = moles * molar mass

Mass Cu(NO3)2 = 0.433 moles * 187.56 g/mol

Mass Cu(NO3)2 = 81.2 grams

Step 8: Calculate actual yield

% yield = (actual yield / theoretical yield) * 100 %

87.3 % = (actual yield / 81.2 grams) * 100 %

0.873 = actual yield / 81.2 grams

Actual yield = 0.873 * 81.2 grams

Actual yield = 70.9 grams

your answer would be “dry”

gas

explanation:

smoke is the matter of a gas