Question 9dacf

Jun 2, 2015

You'd need 58.8 g of phosphoric acid to get that much pyrophosphoric acid.

Pyrophosphoric acid, ${H}_{4} {P}_{2} {O}_{7}$, is obtained by joining two molecules of phosphoric acid, ${H}_{3} P {O}_{4}$, together via condensation, which results in loss of water.

The balanced chemical equation for your reaction looks like this

$\textcolor{red}{2} {H}_{3} P {O}_{4} \to {H}_{4} {P}_{2} {O}_{7} + {H}_{2} O$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between phosphoric acid and pyrophosphoric acid. This means that, regardless of how many moles of the latter were produced, the reaction used up twice as many moles of the former.

Use pyrophosphoric acid's molar mass to determine how many moles were produced

53.4cancel("g") * ("1 mole"H_4P_2O_7)/(178.0cancel("g")) = "0.30 moles" ${H}_{4} {P}_{2} {O}_{7}$

This means that the number of moles of phosphoric acid that reacted was

0.30cancel("moles"H_4P_2O_7) * (color(red)(2)" moles"H_3PO_4)/(1cancel("mole"H_4P_2O_7)) = "0.60 moles"# ${H}_{3} P {O}_{4}$

Now use phosphoric acid's molar mass to see how many grams would contain this many moles

$0.60 \cancel{\text{moles") * "98.0 g"/(1cancel("mole")) = color(green)("58.8 g}}$ ${H}_{3} P {O}_{4}$