in a reaction of 3Cu(s)+ 8HNO3(aq)�¨ 3Cu(NO3)2(aq) +2NO(g) + 4H2O(l). if 5.58 grams of Cu(no3)2 is made how many grams of nitrogen monoxide is present
To find the grams of nitrogen monoxide (NO) produced in the reaction, we need to use stoichiometry.
First, we need to determine the molar ratio between Cu(NO3)2 and NO. From the balanced equation:
3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
We can see that for every 3 moles of Cu(NO3)2 produced, we get 2 moles of NO. This gives us a molar ratio of 3:2.
Now, let's calculate the number of moles of Cu(NO3)2 using its molar mass. The molar mass of Cu(NO3)2 is:
Cu(NO3)2 = (63.55 g/mol Cu) + 2 * [(14.01 g/mol N) + (3 * 16.00 g/mol O)] = 187.55 g/mol
Given that 5.58 grams of Cu(NO3)2 is produced, we can convert it to moles:
moles of Cu(NO3)2 = mass (g) / molar mass (g/mol)
moles of Cu(NO3)2 = 5.58 g / 187.55 g/mol ≈ 0.0297 mol
Now, we can use the molar ratio to find the moles of NO produced:
moles of NO = moles of Cu(NO3)2 * (2 moles of NO / 3 moles of Cu(NO3)2)
moles of NO = 0.0297 mol * (2/3) ≈ 0.0198 mol
Finally, we can convert the moles of NO to grams using its molar mass. The molar mass of NO is:
NO = 14.01 g/mol N + 16.00 g/mol O = 30.01 g/mol
grams of NO = moles of NO * molar mass (g/mol)
grams of NO = 0.0198 mol * 30.01 g/mol ≈ 0.594 g
Therefore, approximately 0.594 grams of nitrogen monoxide (NO) is produced in the reaction.
mols Cu(NO3)2 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols Cu(NO3)2 to mols NO.
mols Cu(NO3)2 x (2 mols NO/3 mol Cu(NO3)2) = mols Cu(NO3)2 x 2/3 = ?
The grams NO = mols NO x molar mass NO.