Chemistry: 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

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A mixture of 22.44 g of CaBr2 and 16.85 g Na3PO4 - id224421685349542930, 101shelsyroman

Ответ:

AlwaysMarcus5577

12.02.2022

Chemistry

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) :

$\tt=\dfrac{22.44}{199,89}=0.112$

mass of Na₃PO₄ = 16.85 g

mol of Na₃PO₄ (MW=163,94 g/mol) :

$\tt =\dfrac{16.85}{163,94}=0.103$

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

$\tt \dfrac{0.112}{3}\div \dfrac{0.103}{2}=0.0373\div 0.0515\rightarrow CaBr_2~limiting~reactant(smaller~ratio)$

So the excess = Na₃PO₄

mol Na₃PO₄ reacted :

$\tt \dfrac{2}{3}\times 0.112=0.0746$

mol Na₃PO₄ unreacted :

$\tt 0.103-0.0746=0.0284$

Mass Na₃PO₄ :

$\tt 0.0284\times 163.94=4.656~g$

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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

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