how many grams of copper hydroxide can b prepared from 2.7g of copper (II) nitrate and excess sodium hydroxide?

See above.

To determine how many grams of copper hydroxide can be prepared from 2.7g of copper (II) nitrate, you'll need to use stoichiometry, which involves balancing the chemical equation and comparing the ratios of the reactants and products.

First, let's write the 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 1 mole of copper (II) nitrate, Cu(NO3)2, reacts with 2 moles of sodium hydroxide, NaOH, to produce 1 mole of copper hydroxide, Cu(OH)2.

Now, let's calculate the number of moles of copper (II) nitrate:
molar mass of Cu(NO3)2 = atomic mass of Cu + 2 * (atomic mass of N + 3 * atomic mass of O)
= 63.55 + 2 * (14 + 3 * 16)
= 63.55 + 2 * (14 + 48)
= 63.55 + 2 * 62
= 63.55 + 124
= 187.55 g/mol

moles of Cu(NO3)2 = mass of Cu(NO3)2 / molar mass of Cu(NO3)2
= 2.7 g / 187.55 g/mol
= 0.0144 mol

Since the mole ratio between Cu(NO3)2 and Cu(OH)2 is 1:1, the number of moles of copper hydroxide that can be prepared is also 0.0144 mol.

Finally, let's calculate the mass of copper hydroxide:
molar mass of Cu(OH)2 = atomic mass of Cu + 2 * (atomic mass of O + atomic mass of H)
= 63.55 + 2 * (16 + 1)
= 63.55 + 2 * 17
= 63.55 + 34
= 97.55 g/mol

mass of Cu(OH)2 = moles of Cu(OH)2 * molar mass of Cu(OH)2
= 0.0144 mol * 97.55 g/mol
= 1.41 g

Therefore, you can prepare approximately 1.41 grams of copper hydroxide from 2.7 grams of copper(II) nitrate and excess sodium hydroxide.