Question

For this reaction, 31.3 g zinc oxide reacts with 11.1 g water.

zinc oxide (s) + water (l) = zinc hydroxide (aq)

What is the maximum mass of zinc hydroxide that can be formed?

What is the FORMULA for the limiting reagent?

What mass 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)= 31.3 g

use:
number of mol of ZnO,
n = mass of ZnO/molar mass of ZnO
=(31.3 g)/(81.38 g/mol)
= 0.3846 mol

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass(H2O)= 11.1 g

use:
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(11.1 g)/(18.02 g/mol)
= 0.6161 mol
Balanced chemical equation is:
ZnO(s) + H2O(l) -> Zn(OH)2(aq)


1 mol of ZnO reacts with 1 mol of H2O
for 0.3846 mol of ZnO, 0.3846 mol of H2O is required
But we have 0.6161 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)*0.3846
= 0.3846 mol


use:
mass of Zn(OH)2 = number of mol * molar mass
= 0.3846*99.4
= 38.23 g
Answer: 38.2 g

2)
ZnO is limiting reagent
Answer: ZnO

3)
According to balanced equation
mol of H2O reacted = (1/1)* moles of ZnO
= (1/1)*0.3846
= 0.3846 mol
mol of H2O remaining = mol initially present - mol reacted
mol of H2O remaining = 0.6161 - 0.3846
mol of H2O remaining = 0.2315 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
= 0.2315 mol * 18.02 g/mol
= 4.171 g
Answer: 4.17 g