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To start with, let us write down the balanced chemical equation for this problem.
`PCl_5 + 4 H_2O -> H_3PO_4 + 5 HCl`
To get the amount of HCl produced in the reaction, we will start with the amount of the reactants. Since both of the reactants weigh 20.0 grams, we should first look for the limiting reactant.
`20.0 grams PCl_5 * (1 mol e PCl_5)/(208.24 grams) * (5 mol es HCl)/(1 mol e PCl_5)`
= 0.4802 moles HCl
`20.0 grams H_2O * (1 mol e H_2O)/(18.0 grams) * (5 mol es HCl)/(4 mol es H_2O)`
= 1.3889 moles HCl
Since PCl5 produces lesser amount of product, it is the limiting reactant and we will continue on solving the amount of HCl from the moles of HCl produced by PCl5.
`= 0.4802 mol es HCl * (36.46 grams HCl)/(1 mol e HCl)`
= 17.5 grams of HCl will be produced.
**Water is the excess reagent in the reaction and therefore will be the left over.
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