10% sulfuric acid solution m = 600 grams was poured with 45 milliliters of ammonia.

10% sulfuric acid solution m = 600 grams was poured with 45 milliliters of ammonia. Calculate the mass of the salt formed.

Given:
ω (H2SO4) = 10%
m solution (H2SO4) = 600 g
V (NH3) = 45 ml = 0.045 l

To find:
m (salt) -?

Decision:
1) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 10% * 600/100% = 60 g;
2) n (H2SO4) = m (H2SO4) / M (H2SO4) = 60/98 = 0.6122 mol;
3) n (NH3) = V (NH3) / Vm = 0.045 / 22.4 = 0.002 mol;
4) H2SO4 + 2NH3 => (NH4) 2SO4;
5) n ((NH4) 2SO4) = n (NH3) / 2 = 0.002 / 2 = 0.001 mol;
6) m ((NH4) 2SO4) = n ((NH4) 2SO4) * M ((NH4) 2SO4) = 0.001 * 132 = 0.132 g.

Answer: The mass of (NH4) 2SO4 is 0.132 g.