Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sul

Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.90 g of sulfuric acid and 5.90g of lead(II) acetate are mixed
Caculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.90 g of sulfuric acid and 5.90g of lead(II) acetate are mixed
Caculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete
Caculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete

Explanation / Answer

H2SO4 + Pb(CH3COO)2 —> PbSO4 + 2 CH3COOH

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)= 5.9 g

use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(5.9 g)/(98.086 g/mol)
= 6.015*10^-2 mol

Molar mass of Pb(CH3COO)2,
MM = 1*MM(Pb) + 4*MM(C) + 6*MM(H) + 4*MM(O)
= 1*207.2 + 4*12.01 + 6*1.008 + 4*16.0
= 325.288 g/mol


mass(Pb(CH3COO)2)= 5.9 g

use:
number of mol of Pb(CH3COO)2,
n = mass of Pb(CH3COO)2/molar mass of Pb(CH3COO)2
=(5.9 g)/(325.288 g/mol)
= 1.814*10^-2 mol
Balanced chemical equation is:
H2SO4 + Pb(CH3COO)2 ---> PbSO4 + 2 CH3COOH


1 mol of H2SO4 reacts with 1 mol of Pb(CH3COO)2
for 6.015*10^-2 mol of H2SO4, 6.015*10^-2 mol of Pb(CH3COO)2 is required
But we have 1.814*10^-2 mol of Pb(CH3COO)2

so, Pb(CH3COO)2 is limiting reagent
we will use Pb(CH3COO)2 in further calculation


Molar mass of PbSO4,
MM = 1*MM(Pb) + 1*MM(S) + 4*MM(O)
= 1*207.2 + 1*32.07 + 4*16.0
= 303.27 g/mol

According to balanced equation
mol of PbSO4 formed = (1/1)* moles of Pb(CH3COO)2
= (1/1)*1.814*10^-2
= 1.814*10^-2 mol


use:
mass of PbSO4 = number of mol * molar mass
= 1.814*10^-2*3.033*10^2
= 5.50 g
Answer: 5.50 g

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