Question

For each of the following precipitation reactions, calculate how many grams of the first reactant are necessary to completely react with 17.9 g of the second reactant.

Na2CO3(aq)+CuCl2(aq)→CuCO3(s)+2NaCl(aq)

Answer 1

The balanced chemical equation for the reaction is:

Na2CO3(aq) + CuCl2(aq) → CuCO3(s) + 2NaCl(aq)

The molar mass of CuCl2 is 134.45 g/mol, so 17.9 g is equivalent to:

17.9 g / 134.45 g/mol = 0.133 mol of CuCl2

From the balanced equation, we know that 1 mole of CuCl2 reacts with 1 mole of Na2CO3. Therefore, we need 0.133 moles of Na2CO3 to react with 0.133 moles of CuCl2.

The molar mass of Na2CO3 is 105.99 g/mol, so we need:

0.133 mol × 105.99 g/mol = 14.12 g of Na2CO3 to react completely with 17.9 g of CuCl2.