Question

question 13 of 24
if 70 moles of naoh combine with 27 moles of h₃po₄, how many moles of h₂o can form? what is the limiting reagent?
3naoh(aq) + h₃po₄(aq) → 3h₂o(l) + na₃po₄(aq)
step 1: show how you would determine the number of moles of h₂o that could form from each starting material.
70 moles naoh × (\frac{}{}) =? moles h₂o
27 moles h₃po₄ × (\frac{}{}) =? moles h₂o
answer bank
3 moles naoh 1 mole h₃po₄
3 moles h₂o 1 mole na₃po₄
step 2: now, calculate the number of moles of h₂o that could form from each starting material.
from naoh: moles h₂o
from h₃po₄: moles h₂o
step 3: identify the limiting reagent.
naoh
h₃po₄

Explanation:

Step1: Set up mole - ratio calculations

For the reaction $3NaOH(aq)+H_3PO_4(aq)
ightarrow3H_2O(l)+Na_3PO_4(aq)$, the mole - ratio of $NaOH$ to $H_2O$ is $\frac{3\ moles\ H_2O}{3\ moles\ NaOH}$, and the mole - ratio of $H_3PO_4$ to $H_2O$ is $\frac{3\ moles\ H_2O}{1\ mole\ H_3PO_4}$. So, $70\ moles\ NaOH\times\frac{3\ moles\ H_2O}{3\ moles\ NaOH}$ and $27\ moles\ H_3PO_4\times\frac{3\ moles\ H_2O}{1\ mole\ H_3PO_4}$.

Step2: Calculate moles of $H_2O$ from each reactant

From $NaOH$: $70\ moles\ NaOH\times\frac{3\ moles\ H_2O}{3\ moles\ NaOH}=70\ moles\ H_2O$.
From $H_3PO_4$: $27\ moles\ H_3PO_4\times\frac{3\ moles\ H_2O}{1\ mole\ H_3PO_4}=81\ moles\ H_2O$.

Step3: Identify the limiting reagent

The limiting reagent is the reactant that produces the least amount of product. Since $NaOH$ produces 70 moles of $H_2O$ and $H_3PO_4$ produces 81 moles of $H_2O$, $NaOH$ is the limiting reagent.

Answer:

from $NaOH$: 70 moles $H_2O$
from $H_3PO_4$: 81 moles $H_2O$
limiting reagent: $NaOH$