Question

For the following reaction, 25.8 grams of zinc oxide are allowed to react with 10.4 grams of water. zinc oxide (s) + water (1) zinc hydroxide (aq) What is the maximum amount of zinc hydroxide that can be formed? grams What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete?

Answer 1

1)

Molar mass of ZnO,

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

= 1*65.38 + 1*16.0

= 81.38 g/mol

mass(ZnO)= 25.8 g

use:

number of mol of ZnO,

n = mass of ZnO/molar mass of ZnO

=(25.8 g)/(81.38 g/mol)

= 0.317 mol

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 10.4 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(10.4 g)/(18.02 g/mol)

= 0.5773 mol

Balanced chemical equation is:

ZnO + H2O ---> Zn(OH)2

1 mol of ZnO reacts with 1 mol of H2O

for 0.317 mol of ZnO, 0.317 mol of H2O is required

But we have 0.5773 mol of H2O

so, ZnO is limiting reagent

we will use ZnO in further calculation

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

According to balanced equation

mol of Zn(OH)2 formed = (1/1)* moles of ZnO

= (1/1)*0.317

= 0.317 mol

use:

mass of Zn(OH)2 = number of mol * molar mass

= 0.317*99.4

= 31.51 g

Answer: 31.5 g

2)

ZnO is limiting reagent

Answer: ZnO

3)

According to balanced equation

mol of H2O reacted = (1/1)* moles of ZnO

= (1/1)*0.317

= 0.317 mol

mol of H2O remaining = mol initially present - mol reacted

mol of H2O remaining = 0.5773 - 0.317

mol of H2O remaining = 0.2602 mol

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

use:

mass of H2O,

m = number of mol * molar mass

= 0.2602 mol * 18.02 g/mol

= 4.688 g

Answer: 4.69 g