Two solutions were mixed, one of which contains 53 g of sodium carbonate, and the other – 152 g of sulfuric acid.

Two solutions were mixed, one of which contains 53 g of sodium carbonate, and the other – 152 g of sulfuric acid. What substance and in what quantity will remain in excess?

Given:
m (Na2CO3) = 53 g
m (H2SO4) = 152 g

To find:
m (remainder) -?

Decision:
1) Na2CO3 + H2SO4 => Na2SO4 + CO2 + H2O;
2) M (Na2CO3) = 106 g / mol;
3) n (Na2CO3) = m (Na2CO3) / M (Na2CO3) = 53/106 = 0.5 mol;
4) M (H2SO4) = 98 g / mol;
5) n (H2SO4) = m (H2SO4) / M (H2SO4) = 152/98 = 1.6 mol;
6) n react. (H2SO4) = n (Na2CO3) = 0.5 mol;
7) n rest. (H2SO4) = n (H2SO4) – n re. (H2SO4) = 1.6 – 0.5 = 1.1 mol;
8) m rest. (H2SO4) = n rest. (H2SO4) * M (H2SO4) = 1.1 * 98 = 107.8 g.

Answer: The mass of the remaining H2SO4 is 107.8 g.