Respuesta :
Answer:
a. 81.23g
b.52.23g
c. 70.91g
Explanation:
[tex]3Cu_{(g)} + 8HNO_{3}[/tex] ⇒ [tex]3Cu(NO_{3}) _{2} + 4H_{2} O + 2NO[/tex]
Cu= 63.5g/mol, H=1g/mol, N=14g/mol, O=16g/mol
3 moles of Cu=190.5g/mol
8 moles of HNO[tex]_{3}[/tex]=504.08g/mol
3 moles of [tex]3Cu(NO_{3}) _{2}[/tex]=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
[tex]\frac{27.5 * 562.71}{190.5}[/tex]=81.23g
b. 190.5 g of Cu will react with 504.08g of HNO3
27.5g of CU will react with
[tex]\frac{27.5 * 504.08}{190.5}[/tex]= 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
[tex]\frac{x}{81.23} * 100=87.3[/tex]
[tex]\frac{100x}{81.23} = 87.3[/tex]
[tex]x=\frac{87.3 * 81.23}{100}[/tex]
x=70.91g
Answer:
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
