Answer:
[tex]89.4\text{ g of H}_3PO_4[/tex]
Explanation:
Here, we want to get the mass of H3PO4 produced
We start by getting the balanced equation of reaction:
[tex]PCl_5\text{ + 4H}_2O\text{ }\rightarrow\text{ 5HCl + H}_3PO_4[/tex]
The mass of H3PO4 produced would be based on the limiting reactant
The limiting reactant here is the one that would produce less amount of the product H3PO4
Firstly, let us get the number of moles of H3PO4 produced by each of the reactants
To get this, we start by getting the number of moles of each that reacted
We get this by dividing the masses by the molar masses
For water, we have it that the molar mass is 18 g/mol
Thus, the number of moles would be:
[tex]\frac{94}{18}\text{ = 5.2 moles}[/tex]
From the balanced equation of reaction:
4 moles of water produced 1 mole of H3PO4
5.2 moles of water will produce x moles of H3PO4
Thus:
[tex]\begin{gathered} x\text{ }\times4\text{ = 5.2}\times1 \\ x\text{ = }\frac{5.2}{4}\text{ = 1.3 mole} \end{gathered}[/tex]
For PCl5
The molar mass is 208 g/mol
Thus,we have the number of moles as:
[tex]\frac{186}{208}\text{ = 0.894 mol}[/tex]
From the equation of reaction:
1 mol of PCl5 produced 1 mol of H3PO4
That means 0.894 mol of PCl5 produced 0.894 mol of H3PO4
Since PCl5 produced a lesser amount of moles of water, that means, it is the limiting reactant
To get the mass of H3PO4 produced, we multiply this number of moles by
the molar mass of H3PO4
The molar mass of H3PO4 is 100 g/mol
Thus, we have the mass of H3PO4 produced as:
[tex]0.894\text{ }\times\text{ 100 = 89.4 g}[/tex]
