Question

#11.) When iron rusts in air, iron(III) oxide is produced. How many moles of oxygen react with 77.4 mol of iron in the rusting reaction?
4Fe(s) + 3O2(g) → 2Fe2O3(s)

Select one:
a. 58
b. 100
c. 77
d. 120

#13.) For the reaction CH4 + 2O2 → CO2 + 2H2O, how many moles of carbon dioxide are produced from the combustion of 161.0 g of methane?

Select one:
a. 10.03
b. 20.06
c. 2584
d. 3.351

Answer 1

Answer 1) Option A) 58.05

In the given reaction of iron forming rust when reacts with the oxygen.

$4 Fe_{(s)} + 3O_{2}_{(g} ----> 2Fe_{2}O_{3}_{(s)}$

We can clearly see that, 4 moles of iron reacts with 3 moles of oxygen to give 2 moles of iron oxide.

So 4 Fe : 3 O and 77.4 moles of Fe : x moles of O

(3 X 77.4) / 4 = 58.05

So when we solve we get x as 58.05.

Hence the no. of moles of oxygen will be 58.

Answer 2) Option A) 10.03

The number of moles of carbon dioxide produced when 161.0 g of methane undergoes combustion will be 10.03

as we know the molar mass of methane is 16.043 g

As we can see in the reaction the mole ratio is 1:1;

1 mole of methane produces 1 mole of carbondioxide.

So, 161 g / 16.043 g = 10.03 moles of Carbon dioxide.