Answer:
a) <u>26.67 moles HNO3 </u>
b) <u>0.33 moles NO</u>
c) <u>0.40 moles NO is produced</u>
d)<u>.157 moles Cu</u>
e) <u>0.105 moles NO</u>
f) <u>26.4 grams HNO3</u>
g) <u>Cu is in excess</u>
h) <u>2.41 grams Cu remain</u>
i) <u>2.37 grams NO</u>
Explanation:
Step 1: Data given
Molar mass of Cu = 63.55 g/mol
Molar mass of HNO3 = 63.01 g/mol
Molar mass of Cu(NO3)2 = 187.56 g/mol
Molar mass of NO = 30.01 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: The balanced equation
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
a) How many moles of HNO3 will react with 10 moles of Cu?
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
For 10 moles Cu we need 8/3 *10 = <u>26.67 moles HNO3 </u>
b) How many moles of NO will form if 0.50 moles of Cu reacts?
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
For 0.50 moles Cu we'll have 2/3 *0.50 = <u>0.33 moles NO</u>
c) If 0.80 moles of H2O forms, how much NO must also form?
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
If 0.80 moles H2O is produced, 0.80/2 = <u>0.40 moles NO is produced</u>
d) How many moles of Cu are in 10.0 grams of Cu?
Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles
In 10.0 grams Cu we have <u>0.157 moles Cu</u>
e) If 10.0 g of Cu reacts, how many moles of NO will form?
10.0 grams Cu = 0.157 moles
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
For 0.157 moles Cu we'll have 2/3 * 0.157 = <u>0.105 moles NO</u>
f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?
10.0 grams Cu = 0.157 moles
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
For 0.157 moles Cu we'll need 0.419 moles HNO3
This is 0.419 moles * 63.01 g/mol = <u>26.4 grams HNO3</u>
g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?
Moles Cu = 0.157 moles
Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
The limiting reactant is HNO3. It will completely be consumed (0.317 moles). <u>Cu is in excess.</u> There will react 3/8 * 0.317 = 0.119 moles Cu
There will remain 0.157 - 0.119 = 0.038 moles
h) How many grams of the excess substance will be left over?
There will react 3/8 * 0.317 = 0.119 moles Cu
There will remain 0.157 - 0.119 = 0.038 moles
This is 0.038 moles * 63.55 g/mol = 2.41 grams
i) How many grams of NO will form in the reaction described in part g?
For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O
For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO
This is 0.079 mol * 30.01 g/mol =<u> 2.37 grams NO</u>
