What mass of MnCl2 will be obtained by the interaction of 25 grams of MnO with hydrochloric acid?

1. Let’s compose the equation of interaction of manganese oxide with hydrochloric acid:
MnO + 2HCl = MnCl2 + H2O;
2.Calculate the chemical amount of manganese oxide:
n (MnO) = m (MnO): M (MnO);
M (MnO) = 55 + 16 = 71 g / mol;
n (MnO) = 25: 71 = 0.3521 mol;
3. Determine the amount and calculate the mass of the manganese chloride obtained:
n (MnCl2) = n (MnO) = 0.3521 mol;
m (MnCl2) = n (MnCl2) * M (MnCl2);
M (MnCl2) = 55 + 35.5 * 2 = 126 g / mol;
m (MnCl2) = 0.3521 * 126 = 44.36 g.
Answer: 44.36 g.