From the stoichiometry of the reaction, the mass of barium nitrate produced is 54.9 g.
The term stoichiometry refers to mass - volume relationships. Stoichiometry can be used to calculate the amount, mass or volume of reactants and products from the balanced reaction equation.
The equation of the reaction is written as follows;
BaCO3 + 2HNO3 ------> Ba(NO3)2 + CO2 + H2O
Number of moles of BaCO3 = 55 g/197.34 g/mol = 0.28 moles
Number of moles of HNO3 = 26 g/63.01 g/mol = 0.41 moles
From the reaction equation;
1 mole of BaCO3 reacts with 2 moles of HNO3
0.28 moles of BaCO3 reacts with 0.28 moles × 2 moles/1 mole = 0.56 moles
There is not enough HNO3 hence it is the limiting reactant.
Number of moles of Ba(NO3)2 produced is obtained from;
2 moles of HNO3 yields 1 mole of Ba(NO3)2
0.41 moles of HNO3 yields 0.41 moles × 1 mole/2 moles
= 0.21 moles of Ba(NO3)2
Mass of Ba(NO3)2 = 0.21 moles × 261.337 g/mol = 54.9 g
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