Question

Please show work

Aqueous sulfuric acid (HI,SO) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na, SO) and liquid water (HO). Suppose 26. g of sulfuric acid is mixed with 7.28 g of sodium hydroacide. 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. Is . X 5 ?

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)= 26.0 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(26 g)/(98.09 g/mol)

= 0.2651 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)= 7.28 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(7.28 g)/(40 g/mol)

= 0.182 mol

Balanced chemical equation is:

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

1 mol of H2SO4 reacts with 2 mol of NaOH

for 0.2651 mol of H2SO4, 0.5301 mol of NaOH is required

But we have 0.182 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)*0.182

= 9.1*10^-2 mol

use:

mass of Na2SO4 = number of mol * molar mass

= 9.1*10^-2*1.421*10^2

= 12.93 g

Answer: 13 g