When copperIl) nitrate reacts with sodium hydroxide, copperIl) hydroxide is produced. How many grams of Cu(OH, can be prepared from 12.7 g of Cu(NO,2 and excess NaOH?
To determine the amount of Cu(OH)2 produced, we need to first write out the balanced chemical equation for the reaction between copper(II) nitrate (Cu(NO3)2) and sodium hydroxide (NaOH):
Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3
From the balanced equation, we can see that 1 mole of Cu(NO3)2 produces 1 mole of Cu(OH)2.
First, we need to find 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 + (3 * 16.00) = 187.55 g/mol
Next, we calculate the number of moles of Cu(NO3)2 in 12.7 g:
Number of moles = Mass / Molar mass
Number of moles = 12.7 g / 187.55 g/mol ≈ 0.0677 mol
Since 1 mole of Cu(NO3)2 produces 1 mole of Cu(OH)2, the number of moles of Cu(OH)2 produced is also 0.0677 mol.
Now, we need to find the molar mass of Cu(OH)2:
Cu = 63.55 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of Cu(OH)2 = 63.55 + 2(16.00) + 2(1.01) = 97.57 g/mol
Finally, we can calculate the mass of Cu(OH)2 produced:
Mass = Number of moles * Molar mass
Mass = 0.0677 mol * 97.57 g/mol ≈ 6.60 g
Therefore, approximately 6.60 grams of Cu(OH)2 can be produced from 12.7 grams of Cu(NO3)2 and excess NaOH.