calculate the amount of aluminum nitrate produced from 50.00 grams of copper II nitrate?

Question

calculate the amount of aluminum nitrate produced from 50.00 grams of copper II nitrate?

3 Cu(NO₃)₂ + 2 AlCl₃ → 3 CuCl₂ + 2 Al(NO₃)₃

calculate the amount of HNO3 required to complete react with 50grams of copper II hydroxide?

2 HNO₃ + Cu(OH)₂ → Cu(NO₃)₂ + 2 H₂O

if 50.00 grams H2SO4 react with 5o.00 grams Cu, how many grams of hydrogen can be produced? which is the limiting reactant? how much excess reactants will be left over?

Cu⁰ + 2 H₂S^VIO₄ → S^IVO₂ + 2 H₂O + Cu^IIS^VIO₄

please help me answer these thank you

Answer 1

Didn't you ask this once before, and didn't I post an answer?

3Cu(NO₃)₂ + 2AlCl₃ ===> 3CuCl₂ + 2Al(NO₃)₃

Assuming aluminum chloride is NOT limiting, then amount of aluminum nitrate will depend only on Cu(NO₃)₂

50.00 g Cu(NO₃)₂ x 1 mole/187.56 g x 2 mole Al(NO₃)₃/3 moles Cu(NO₃)₂ x 212.996 g/mole Al(NO₃)₃ = 37.85 g of aluminum nitrate.

2HNO3 + Cu(OH)2 ===> Cu(NO3)2 + 2H2O

50 g Cu(OH)2 x 1 mole/97.561 g x 2 mole HNO3/mole Cu(OH)2 x 63.01 g/mol HNO3 = 64.58 g HNO3 (this is to 4 significant figures even though the question uses 50 g copper(II) hydroxide which has only 1 sig. fig. Since the other 2 questions use 4 sig. figs., I assume this question was supposed to be 50.00 g).

Cu + 2H₂SO₄ ===> SO₂ + 2H₂O + CuSO₄

moles Cu present = 50.00 g x 1 mol/63.55 g = 0.7868 moles Cu

moles H₂SO₄ = 50.00 g x 1 mol/98.079 g = 0.5098 moles H₂SO₄

H₂SO₄ is the limiting reactant because it takes 2 moles per every mole of Cu, and there isn't enough.

The question asks for the grams of hydrogen, but hydrogen IS NOT produced in this reaction. The Cu is a relatively inactive metal and will not produce hydrogen gas in this reaction with H₂SO₄.