b0.795 g fon . Calculate the mass of excess reactant remaining after 4.50 mol of LiHCO3...

Question

Question

Answer 1

Answer #1

Balanced chemical equation is:

2 LiHCO3 + H2SO4 ---> Li2SO4 +

2 mol of LiHCO3 reacts with 1 mol of H2SO4

for 4.5 mol of LiHCO3, 2.25 mol of H2SO4 is required

But we have 8.5 mol of H2SO4

so, LiHCO3 is limiting reagent

According to balanced equation

mol of H2SO4 reacted = (1/2)* moles of LiHCO3

= (1/2)*4.5

= 2.25 mol

mol of H2SO4 remaining = mol initially present - mol reacted

mol of H2SO4 remaining = 8.5 - 2.25

mol of H2SO4 remaining = 6.25 mol

Molar mass of H2SO4 = 98.08 g/mol

use:

mass of H2SO4,

m = number of mol * molar mass

= 6.25 mol * 98.08 g/mol

= 613 g

Answer: d

Answer 2

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