Question

if 35.0 ml of 0.150M H3PO4 is combined with 95.0 ml of .150M NaOH how many grams of water are theoretically produced by the reaction?

Answer 1

volume of H3PO4, V = 35 mL

= 3.5*10^-2 L

use:

number of mol in H3PO4,

n = Molarity * Volume

= 0.15*3.5*10^-2

= 5.25*10^-3 mol

volume of NaOH, V = 95 mL

= 9.5*10^-2 L

use:

number of mol in NaOH,

n = Molarity * Volume

= 0.15*9.5*10^-2

= 1.425*10^-2 mol

Balanced chemical equation is:

H3PO4 + 3 NaOH ---> 3 H2O + Na3PO4

1 mol of H3PO4 reacts with 3 mol of NaOH

for 5.25*10^-3 mol of H3PO4, 1.575*10^-2 mol of NaOH is required

But we have 1.425*10^-2 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (3/3)* moles of NaOH

= (3/3)*1.425*10^-2

= 1.425*10^-2 mol

use:

mass of H2O = number of mol * molar mass

= 1.425*10^-2*18.02

= 0.2567 g

Answer: 0.257 g