Question

A mixture of 22.44 g of CaBr2 and 16.85 g Na3PO4 is used in the following reaction. Determine the mass of the excess reactant left unreacted.

3 CaBr2 + 2 Na3PO4 --> Ca3(PO4)2 + 6NaBr

Answer 1

The mass of the excess reactant left unreacted : 4.656 g

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products

Reaction

3 CaBr₂ + 2 Na₃PO₄ --> Ca₃(PO₄)₂ + 6NaBr

mass of CaBr₂ = 22.44 g

mol of CaBr₂(MW=199,89 g/mol) :

`\tt=(22.44)/(199,89)=0.112`

mass of Na₃PO₄ = 16.85 g

mol of Na₃PO₄ (MW=163,94 g/mol) :

`\tt =(16.85)/(163,94)=0.103`

mol ratio of reactants (to find limiring reactant): CaBr₂ : Na₃PO₄ =

`\tt (0.112)/(3)/ (0.103)/(2)=0.0373/ 0.0515\rightarrow CaBr_2~limiting~reactant(smaller~ratio)`

So the excess = Na₃PO₄

mol Na₃PO₄ reacted :

`\tt (2)/(3)* 0.112=0.0746`

mol Na₃PO₄ unreacted :

`\tt 0.103-0.0746=0.0284`

Mass Na₃PO₄ :

`\tt 0.0284* 163.94=4.656~g`