#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) Option A) 58.05

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

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.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2