copper metal reacts with nitric acid. assume that the reaction is 3Cu(s) + 8HNO3(aq) ——> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l). if 5.58 grams of Cu(NO3)2 is obtained, how many grams of nitrogen monoxide would have formed
how many moles of Cu(NO3)2 in 5.58g?
2 moles 2NO are produced for every 3 moles Cu(NO3)2.
To find the number of grams of nitrogen monoxide (NO) formed, we first need to determine the molar ratio between Cu(NO3)2 and NO in the balanced chemical equation.
From the balanced equation:
3 Cu(NO3)2 (aq) + 2 NO(g) + 4 H2O(l)
We can see that the ratio between Cu(NO3)2 and NO is 3:2.
To calculate the number of moles of Cu(NO3)2, we can use the formula:
moles = mass / molar mass
The molar mass of Cu(NO3)2 can be calculated by adding up the atomic masses of its components:
Cu(NO3)2:
- Cu: 63.55 g/mol
- N: 14.01 g/mol (from nitrate, NO3)
- O: 16.00 g/mol (from nitrate, NO3)
The molar mass of Cu(NO3)2 is:
63.55 g/mol (Cu) + 2 * (14.01 g/mol (N) + 3 * 16.00 g/mol (O))
= 63.55 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 63.55 g/mol + 2 * (62.01 g/mol)
= 63.55 g/mol + 124.02 g/mol
= 187.57 g/mol
Now, we can calculate the number of moles of Cu(NO3)2:
moles = mass / molar mass
moles = 5.58 g / 187.57 g/mol
moles = 0.0297 mol
Since the ratio between Cu(NO3)2 and NO is 3:2, we can calculate the number of moles of NO formed:
moles of NO = (2/3) * moles of Cu(NO3)2
moles of NO = (2/3) * 0.0297 mol
moles of NO = 0.0198 mol
To determine the mass of NO formed, we can use its molar mass:
Molar mass of NO = 14.01 g/mol (N) + 16.00 g/mol (O)
Molar mass of NO = 30.01 g/mol
Now, we can calculate the mass of NO:
mass = moles * molar mass
mass = 0.0198 mol * 30.01 g/mol
mass = 0.594 g
Therefore, approximately 0.594 grams of nitrogen monoxide (NO) would have been formed.
To find the number of grams of nitrogen monoxide (NO) that would have formed in the reaction, we can use stoichiometry.
First, let's calculate the molar mass of Cu(NO3)2:
Cu: 63.55 g/mol
N: 14.01 g/mol
O: 16.00 g/mol
Molar mass of Cu(NO3)2 = (63.55 g/mol) + 2 * [(14.01 g/mol) + 3 * (16.00 g/mol)]
= 63.55 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 63.55 g/mol + 2 * (62.01 g/mol)
= 63.55 g/mol + 124.02 g/mol
= 187.57 g/mol
Now, let's set up a proportion using the balanced equation:
3 moles of Cu(NO3)2 produce 2 moles of NO
Using the molar mass of Cu(NO3)2, we can calculate the number of moles:
5.58 grams of Cu(NO3)2 * (1 mol / 187.57 g) = 0.0297 moles of Cu(NO3)2
Now, we can use stoichiometry to find the number of moles of NO:
0.0297 moles of Cu(NO3)2 * (2 moles of NO / 3 moles of Cu(NO3)2) = 0.0198 moles of NO
Finally, we can calculate the grams of NO:
0.0198 moles of NO * (30.01 g/mol) = 0.594 grams of NO
Therefore, approximately 0.594 grams of nitrogen monoxide (NO) would have formed.