Question

References In progress Use the References to access important values if needed for this question. For the following reaction, 35.2 grams of sulfuric acid are allowed to react with 40.6 grams of zinc hydroxide. sulfuric acid (aq) + zinc hydroxide (s) — zinc sulfate (aq) + water (1) What is the maximum amount of zinc sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer

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)= 35.2 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(35.2 g)/(98.09 g/mol)

= 0.3589 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)= 40.6 g

use:

number of mol of Zn(OH)2,

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

=(40.6 g)/(99.4 g/mol)

= 0.4085 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 0.3589 mol of H2SO4, 0.3589 mol of Zn(OH)2 is required

But we have 0.4085 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)*0.3589

= 0.3589 mol

use:

mass of ZnSO4 = number of mol * molar mass

= 0.3589*1.614*10^2

= 57.94 g

Answer: 57.9 g

2)

H2SO4 is limiting reagent

Answer: H2SO4

3)

According to balanced equation

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

= (1/1)*0.3589

= 0.3589 mol

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

mol of Zn(OH)2 remaining = 0.4085 - 0.3589

mol of Zn(OH)2 remaining = 4.96*10^-2 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

use:

mass of Zn(OH)2,

m = number of mol * molar mass

= 4.96*10^-2 mol * 99.4 g/mol

= 4.93 g

Answer: 4.93 g