Sam G.

asked • 03/24/21

Limiting Reactions

5. In the reaction BaCO3 + 2HNO3 ---> Ba (NO3)2 + CO2 + H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3?

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Sidney P. answered • 03/25/21

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With 55g BaCO3 * (1 mol BaCO3 / 197.3g BaCO3) * (1 mol Ba(NO3)2 / 1 mol BaCO3) = 0.279 mol Ba(NO3)2.

With 26g HNO3 * (1 mol HNO3 / 63.0g HNO3) * (1 mol Ba(NO3)2 / 2 mol HNO3) = 0.206 mol Ba(NO3)2. Thus HNO3 is the limiting reactant.

Mass of Ba(NO3)2 = (0.206 mol Ba(NO3)2) * (261.3g Ba(NO3)2 / 1 mol Ba(NO3)2) = 53.9 g Ba(NO3)2.

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