Question

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?​

Answer 1

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

Step-by-step 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

Answer 2

a. 81.23g

b.52.23g

c. 70.91g

Step-by-step explanation:

`3Cu_((g)) + 8HNO_(3)` ⇒
`3Cu(NO_(3)) _(2) + 4H_(2) O + 2NO`

Cu= 63.5g/mol, H=1g/mol, N=14g/mol, O=16g/mol

3 moles of Cu=190.5g/mol

8 moles of HNO
`_(3)`=504.08g/mol

3 moles of
`3Cu(NO_(3)) _(2)`=562.71g/mol

a. Cu is the limiting reagent. A limiting reagent is the reactant that determines how much of the products are made, usually the reactant in smaller quantity. Using the limiting reagent for calculations;

190.5g of Cu will yield 562.71g of Cu(NO3)2

27.5g of Cu will need

`(27.5 * 562.71)/(190.5)`=81.23g

b. 190.5 g of Cu will react with 504.08g of HNO3

27.5g of CU will react with

`(27.5 * 504.08)/(190.5)`= 72.77g

grams of excess reactant=125g-72.77g=52.23g

c. let x be the actual yield of CU(NO3)2, from answer a, the theoretical produce

`(x)/(81.23) * 100=87.3`

`(100x)/(81.23) = 87.3`

`x=(87.3 * 81.23)/(100)`

x=70.91g