Question

Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O . Suppose 92. g of sulfuric acid is mixed with 39.9 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 92.0 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(92 g)/(98.09 g/mol)

= 0.938 mol

Molar mass of NaOH,

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

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 39.9 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(39.9 g)/(40 g/mol)

= 0.9975 mol

Balanced chemical equation is:

H2SO4 + 2 NaOH ---> Na2SO4 + 2 H2O

1 mol of H2SO4 reacts with 2 mol of NaOH

for 0.938 mol of H2SO4, 1.876 mol of NaOH is required

But we have 0.9975 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of Na2SO4,

MM = 2*MM(Na) + 1*MM(S) + 4*MM(O)

= 2*22.99 + 1*32.07 + 4*16.0

= 142.05 g/mol

According to balanced equation

mol of Na2SO4 formed = (1/2)* moles of NaOH

= (1/2)*0.9975

= 0.4988 mol

use:

mass of Na2SO4 = number of mol * molar mass

= 0.4988*1.421*10^2

= 70.85 g

Answer: 71 g