How many grams of magnesium carbonate can be obtained from 500 grams of magnesium chloride and sodium carbonate.

Given:
m (MgCl2) = 500 g

To find:
m (MgCO3) -?

Decision:
1) MgCl2 + Na2CO3 => MgCO3 ↓ + 2NaCl;
2) M (MgCl2) = Mr (MgCl2) = Ar (Mg) * N (Mg) + Ar (Cl) * N (Cl) = 24 * 1 + 35.5 * 2 = 95 g / mol;
3) n (MgCl2) = m (MgCl2) / M (MgCl2) = 500/95 = 5.26 mol;
4) n (MgCO3) = n (MgCl2) = 5.26 mol;
5) M (MgCO3) = Mr (MgCO3) = Ar (Mg) * N (Mg) + Ar (C) * N (C) + Ar (O) * N (O) = 24 * 1 + 12 * 1 + 16 * 3 = 84 g / mol;
6) m (MgCO3) = n (MgCO3) * M (MgCO3) = 5.26 * 84 = 441.8 g.

Answer: The mass of MgCO3 is 441.8 g.