Question

study guide, please show all work

4. Aqueous sulfuric acid (H2SO.) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO.) and liquid water (H2O). If 5.76 g of sodium sulfate is produced from the reaction of 7.8 g of sulfuric acid and 9.5 g of sodium hydroxide, calculate the percent yield of sodium sulfate. Round your answer to 2 significant figures.

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)= 7.8 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(7.8 g)/(98.09 g/mol)

= 7.952*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)= 9.5 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(9.5 g)/(40 g/mol)

= 0.2375 mol

Balanced chemical equation is:

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

1 mol of H2SO4 reacts with 2 mol of NaOH

for 7.952*10^-2 mol of H2SO4, 0.159 mol of NaOH is required

But we have 0.2375 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)*7.952*10^-2

= 7.952*10^-2 mol

use:

mass of Na2SO4 = number of mol * molar mass

= 7.952*10^-2*1.421*10^2

= 11.3 g

% yield = actual mass*100/theoretical mass

= 5.76*100/11.3

= 51.0 %

Answer: 51.0 %