Question

it is desired to produce 9.32 grams of phosphoric acid by the following reaction. if the percent yield of phosphoric acid is 72.0%, how many grams of water would need to be reacted? diphosphorus pentoxide(s) + water(l)→phosphoric acid(aq) 3 item attempts remaining grams water numeric input field

Explanation:

Step1: Write the balanced chemical equation

$P_2O_5 + 3H_2O
ightarrow 2H_3PO_4$

Step2: Calculate the theoretical yield of $H_3PO_4$

Let the theoretical yield be $x$ grams. Given percent - yield = 72.0% and actual yield = 9.32 g. Percent - yield=$\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$. So, $72.0=\frac{9.32}{x}\times100$. Solving for $x$ gives $x=\frac{9.32\times100}{72.0}\approx12.94$ g.

Step3: Calculate the moles of $H_3PO_4$ in the theoretical yield

The molar mass of $H_3PO_4$ is $M=(3\times1 + 31+4\times16)=98$ g/mol. The moles of $H_3PO_4$, $n=\frac{12.94}{98}\approx0.132$ mol.

Step4: Determine the moles of $H_2O$ from the stoichiometry

From the balanced equation, the mole - ratio of $H_2O$ to $H_3PO_4$ is 3:2. So the moles of $H_2O$, $n_{H_2O}=\frac{3}{2}\times0.132 = 0.198$ mol.

Step5: Calculate the mass of $H_2O$

The molar mass of $H_2O$ is $M_{H_2O}=(2\times1 + 16)=18$ g/mol. The mass of $H_2O$, $m = n_{H_2O}\times M_{H_2O}=0.198\times18 = 3.56$ g.

Answer:

3.56