Respuesta :
Answer:
a) 26.67 moles HNO3
b) 0.33 moles NO
c) 0.40 moles NO is produced
d).157 moles Cu
e) 0.105 moles NO
f) 26.4 grams HNO3
g) Cu is in excess
h) 2.41 grams Cu remain
i) 2.37 grams NO
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 = 26.67 moles HNO3
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 = 0.33 moles NO
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 = 0.40 moles NO is produced
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 0.157 moles Cu
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 = 0.105 moles NO
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 = 26.4 grams HNO3
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). Cu is in excess. 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 = 2.37 grams NO
