In a double replacement reaction copper (II) chloride combines with sodium nitrate. When a student combines 3 grams of copper (II) chloride react with 4 grams of sodium nitrate, how much sodium chloride is produced?

0.0446 moles of NaCl are produced by the reaction

Explanation:

First of all we convert the mass of reactants to moles, to determine the limiting reagent

3 g / 134.45 g/mol =0.0223 moles of CuCl₂

4 g / 85 g/mol = 0.0470 moles of NaNO₃

The reaction is: CuCl₂(aq) + 2NaNO₃(aq) →  2NaCl(aq)  +  Cu(NO₃)₂(aq)

By stoichiomety, 1 mol of chloride reacts with 2 moles of nitrate

Then, 0.0223 moles of chloride will react with (0.0223 . 2) / 1 = 0.0446 moles of nitrate. We have 0.0470 moles, so the nitrate is in excess, and the limiting reactant is the chloride.

2 moles of nitrate react with 1 mol of chloride

Then, 0.0470 moles of nitrate will react with (0.0470 . 1) / 2 = 0.0235 moles of chloride. There is not enough chloride because we have 0.0223 moles and we need 0.0235.

In the reaction, 1 mol of chloride produces 2 mol of sodium chloride

Then, our 0.0223 moles will produce (0.0223 .2 ) / 1 = 0.0446 moles of NaCl

idk what the question is about