Question

If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium phosphate will produced? How many grams of excess reactant will be left over?

Answer 1

Answer:

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

• H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

First we convert the mass of both reactants to moles, using their respective molar masses:

• H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol

• NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

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We calculate the produced moles of Na₃PO₄ using the limiting reactant:

• 3.75 mol NaOH * $\frac{1molNa_3PO_4}{3molNaOH}$ = 1.25 mol Na₃PO₄

Then we convert moles into grams:

• 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g

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We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

• 3.75 mol NaOH * $\frac{1molH_3PO_4}{3molNaOH}$ = 1.25 mol H₃PO₄

We substract that amount from the original amount:

• 1.78 - 1.25 = 0.53 mol H₃PO₄

Finally we convert those remaining moles to grams:

• 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g