Determine the mass of ammonium nitrate, which should be obtained by interaction of 50 liters of ammonia and 630 g of 20% nitric acid.

univerkov | November 25, 2020 | education

NH3 + HNO3 = NH4NO3
n (NH3) = 50 / 22.4 = 2.23214285714 mol
m (HNO3) = 630 * 0.2 = 126 g
n (HNO3) = 126/63 = 2 mol n (NH3)> n (HNO3)
m (NH4NO3) = 0.2 * 80 = 16 g

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