Question

For the following reaction, 31.9 grams of sulfuric acid are allowed to react with 37.2 grams of zinc hydroxide.

sulfuric acid (aq) + zinc hydroxide (s) zinc sulfate (aq) + water (l)

What is the maximum amount of zinc sulfate that can be formed?

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?

Answer 1

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)= 31.9 g

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(31.9 g)/(98.086 g/mol)

= 0.3252 mol

Molar mass of Zn(OH)2,

MM = 1*MM(Zn) + 2*MM(O) + 2*MM(H)

= 1*65.38 + 2*16.0 + 2*1.008

= 99.396 g/mol

mass(Zn(OH)2)= 37.2 g

number of mol of Zn(OH)2,

n = mass of Zn(OH)2/molar mass of Zn(OH)2

=(37.2 g)/(99.396 g/mol)

= 0.3743 mol

Balanced chemical equation is:

H2SO4 + Zn(OH)2 ---> ZnSO4 + 2 H2O

1 mol of H2SO4 reacts with 1 mol of Zn(OH)2

for 0.3252 mol of H2SO4, 0.3252 mol of Zn(OH)2 is required

But we have 0.3743 mol of Zn(OH)2

so, H2SO4 is limiting reagent

we will use H2SO4 in further calculation

Molar mass of ZnSO4,

MM = 1*MM(Zn) + 1*MM(S) + 4*MM(O)

= 1*65.38 + 1*32.07 + 4*16.0

= 161.45 g/mol

According to balanced equation

mol of ZnSO4 formed = (1/1)* moles of H2SO4

= (1/1)*0.3252

= 0.3252 mol

mass of ZnSO4 = number of mol * molar mass

= 0.3252*1.614*10^2

= 52.5 g

Answer: 52.5 g

2)

Answer: H2SO4 is limiting reagent

3)

According to balanced equation

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

= (1/1)*0.3252

= 0.3252 mol

mol of Zn(OH)2 remaining = mol initially present - mol reacted

mol of Zn(OH)2 remaining = 0.3743 - 0.3252

mol of Zn(OH)2 remaining = 0.049 mol

Molar mass of Zn(OH)2,

MM = 1*MM(Zn) + 2*MM(O) + 2*MM(H)

= 1*65.38 + 2*16.0 + 2*1.008

= 99.396 g/mol

mass of Zn(OH)2,

m = number of mol * molar mass

= 4.904*10^-2 mol * 99.396 g/mol

= 4.87 g

Answer: 4.87 g