To determine the grams of copper (II) hydroxide produced, we need to first write a balanced equation for the reaction between copper (II) nitrate and sodium hydroxide:
Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3
From the balanced equation, we can see that the ratio of copper (II) nitrate to copper (II) hydroxide is 1:1. This means that for every 1 mole of copper (II) nitrate, we will obtain 1 mole of copper (II) hydroxide.
To calculate the moles of copper (II) hydroxide produced, we need to convert the given mass of copper (II) nitrate to moles. We will use the molar mass of copper (II) nitrate to perform this conversion.
The molar mass of copper (II) nitrate (Cu(NO3)2) can be calculated as follows:
Cu: 63.55 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 oxygen atoms)
Molar mass of Cu(NO3)2 = (63.55 g/mol) + (2 * (14.01 g/mol)) + (6 * (16.00 g/mol)) = 187.55 g/mol
Now, we can calculate the moles of copper (II) nitrate:
moles = mass / molar mass = 2.7 g / 187.55 g/mol ≈ 0.0144 mol
Since the ratio of copper (II) nitrate to copper (II) hydroxide is 1:1, the moles of copper (II) hydroxide produced will also be 0.0144 mol.
Finally, we can calculate the mass of copper (II) hydroxide using its molar mass.
The molar mass of copper (II) hydroxide (Cu(OH)2) can be calculated as follows:
Cu: 63.55 g/mol
O: 16.00 g/mol (2 oxygen atoms)
H: 1.01 g/mol (2 hydrogen atoms)
Molar mass of Cu(OH)2 = (63.55 g/mol) + (2 * (1.01 g/mol)) + (2 * (16.00 g/mol)) = 97.55 g/mol
Now, we can calculate the mass of copper (II) hydroxide:
mass = moles * molar mass = 0.0144 mol * 97.55 g/mol ≈ 1.41 g (rounded to two decimal places)
Therefore, approximately 1.41 grams of copper (II) hydroxide can be prepared from 2.7 grams of copper (II) nitrate reacting with sodium hydroxide.