Question

Use the Kererences to access important eeded for thi For the following reaction, 31.7 grams of zinc oxide are allowed to react with 10.6 grams of wate zinc oxide() + water(l) — zinc hydroxide(aq) What is the maximum mass of zine 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? grams Submit Answer Retry Entire Group 8 more group attempts remaining

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)= 31.7 g

use:

number of mol of ZnO,

n = mass of ZnO/molar mass of ZnO

=(31.7 g)/(81.38 g/mol)

= 0.3895 mol

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 10.6 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(10.6 g)/(18.02 g/mol)

= 0.5884 mol

Balanced chemical equation is:

ZnO(s) + H2O(l) ---> Zn(OH)2(aq)

1 mol of ZnO reacts with 1 mol of H2O

for 0.3895 mol of ZnO, 0.3895 mol of H2O is required

But we have 0.5884 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)*0.3895

= 0.3895 mol

use:

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

= 0.3895*99.4

= 38.72 g

Answer: 38.7 g

2)

ZnO is limiting reagent

Answer: ZnO

3)

According to balanced equation

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

= (1/1)*0.3895

= 0.3895 mol

mol of H2O remaining = mol initially present - mol reacted

mol of H2O remaining = 0.5884 - 0.3895

mol of H2O remaining = 0.1988 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

= 0.1988 mol * 18.02 g/mol

= 3.582 g

Answer: 3.58 g