Question

Calculate the number of grams of excess reagent remaining when 17.3 grams of H2SO4 is reacted with 43.4 grams of NaOH in the reaction: H2SO4 + 2 NaOH ---> 2H2O + Na2SO4

Answer 1

Molar mass of H2SO4 = 2*MM(H) + 1*MM(S) + 4*MM(O)

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

= 98.086 g/mol

mass of H2SO4 = 17.3 g

we have below equation to be used:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(17.3 g)/(98.086 g/mol)

= 0.1764 mol

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

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

= 39.998 g/mol

mass of NaOH = 43.4 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(43.4 g)/(39.998 g/mol)

= 1.085 mol

we have the Balanced chemical equation as:

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

1 mol of H2SO4 reacts with 2 mol of NaOH

for 0.1764 mol of H2SO4, 0.3528 mol of NaOH is required

But we have 1.085 mol of NaOH

so, H2SO4 is limiting reagent

we will use H2SO4 in further calculation

From balanced chemical reaction, we see that

when 1 mol of H2SO4 reacts, 2 mol of NaOH reacts

mol of NaOH reacted = (2/1)* moles of H2SO4

= (2/1)*0.1764

= 0.3528 mol

mol of NaOH remaining = mol initially present - mol reacted

mol of NaOH remaining = 1.085 - 0.3528

mol of NaOH remaining = 0.7323 mol

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

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

= 39.998 g/mol

we have below equation to be used:

mass of NaOH,

m = number of mol * molar mass

= 0.7323 mol * 39.998 g/mol

= 29.29 g

Answer: 29.3 g