Question

given 5.0 moles of koh and 2.0 moles of h₃po₄, how many moles of k₃po₄ can be prepared? h₃po₄ + 3koh → k₃po₄ + 3h₂o 1.7 moles 1.2 moles 5.0 moles 2.0 moles 3.5 moles

Explanation:

Step1: Determine the limiting reactant

From the balanced equation \(H_3PO_4 + 3KOH
ightarrow K_3PO_4+3H_2O\), the mole ratio of \(H_3PO_4\) to \(KOH\) is \(1:3\).
If we have \(n(H_3PO_4) = 2.0\space mol\), the amount of \(KOH\) required for complete reaction is \(n_{required}(KOH)=3\times n(H_3PO_4)=3\times2.0\space mol = 6.0\space mol\). But we have \(n(KOH) = 5.0\space mol\).
If we use \(KOH\) as the basis, the amount of \(H_3PO_4\) required for complete reaction of \(KOH\) is \(n_{required}(H_3PO_4)=\frac{n(KOH)}{3}=\frac{5.0\space mol}{3}\approx1.7\space mol\). Since \(1.7\space mol<2.0\space mol\), \(KOH\) is the limiting reactant.

Step2: Calculate the moles of \(K_3PO_4\)

From the balanced equation, the mole ratio of \(KOH\) to \(K_3PO_4\) is \(3:1\).
Let \(n(K_3PO_4)\) be the moles of \(K_3PO_4\). Using the mole - ratio \(\frac{n(K_3PO_4)}{n(KOH)}=\frac{1}{3}\), and \(n(KOH) = 5.0\space mol\)
\(n(K_3PO_4)=\frac{n(KOH)}{3}=\frac{5.0\space mol}{3}\approx1.7\space mol\)

Answer:

1.7 moles