Question

Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O . Suppose 2.9 g of sulfuric acid is mixed with 3.58 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)= 2.9 g

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(2.9 g)/(98.086 g/mol)

= 2.957*10^-2 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)= 3.58 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(3.58 g)/(39.998 g/mol)

= 8.95*10^-2 mol

Balanced chemical equation is:

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

1 mol of H2SO4 reacts with 2 mol of NaOH

for 0.0296 mol of H2SO4, 0.0591 mol of NaOH is required

But we have 0.0895 mol of NaOH

so, H2SO4 is limiting reagent

we will use H2SO4 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 = (1/1)* moles of H2SO4

= (1/1)*0.0296

= 0.0296 mol

mass of Na2SO4 = number of mol * molar mass

= 2.957*10^-2*1.421*10^2

= 4.2 g

Answer: 4.2 g