Chem: Finding theoretical yield

How much sodium carbonate is required to react with 20.0g KAuCl4 in the following reaction? What is the theoretical yield of gold (III) hydroxide?

2KAuCl4 + 3Na2CO3 + 3H2O -> 2Au(OH)3 + 6NaCl + 2KCl + 3CO2




Please help! Show your work and explain your steps clearly please!

Sodium carbonate:
20.0 g KAuCl4 * (1 mol KAuCl4 / 377.88 g KAuCl4) * (3 mol Na2CO3 / 2 mol KAuCl4) * (105.9894 g Na2CO3 / 1 mol Na2CO3)
= 8.414528422 g Na2CO3
With significant figures, that is 8.41 g Na2CO3.
(The molar mass of KAuCl4 is 377.88 g/mol. The molar mass of Na2CO3 is 105.9894 g/mol.)

Theoretical yield of gold(III) hydroxide:
20.0 g KAuCl4 * (1 mol KAuCl4 / 377.88 g KAuCl4) * (2 mol Au(OH)3 / 2 mol KAuCl4) * (247.99211 g Au(OH)3 / 1 mol Au(OH)3)
= 13.12544247 g Au(OH)3
With significant figures, that is 13.1 g.
(The molar mass of Au(OH)3 is 247.99211 g/mol.)