Answer to Question #180198 in General Chemistry for hayley samantha farah

Question

Use the following balanced equation to answer each part below:

H3PO4​+3NaOH Na3​PO4​+3H2​O

If you start the reaction with 100. grams of each reactant, which reactant is the limiting reactant, and how many grams of the excess reactant are left over? What mass of sodium phosphate is produced?

The limiting reactant is:

How much of the excess reactant remains?

How many grams of sodium phosphate are produced in this reaction?

Answer 1

Moles of H3PO4 in 100g $=\dfrac{100g}{97.997g/mol} = 1.0205mol$

Moles of NaOH required to completely react with H3PO4;

Mole ratio = 1:3

Moles of NaOH $=\dfrac{3}{1}x1.0205mol = 3.0614mol$

But moles of NaOH in 100g $=\dfrac{100g}{40g/mol}= 2.5 mol$ which is less than required

Thus NaOH is the limiting reagent

Moles of H3PO4 that will react with 2.5mol of NaOH are;

$=\dfrac{1}{3}x2.5mol = 0.8333mol$

Moles of H3PO4 that will remain (excess) = (1.0205-0.8333) = 0.1872 mol

Moles of product will depend on limiting reagent (NaOH)

Mole ratio = 3:1

Moles of Na3PO4 $\dfrac{1}{3}x 2.5 = 0.8333mol$

Mass of Na3PO4 = moles x MM

= 0.8333mol x 163.941g/mol = 136.62g