Question

A) If 30.8 g of O₂ are mixed with 30.8 g of H₂ and the mixture is ignited, what mass of water is produced?

B) Iron is produced from its ore by the reactions:

2C(s)+O₂(g) → 2CO(g)

Fe₂O₃(s)+3CO(g) → 2Fe(s) + 3co₂(g)

How many moles of O₂(g) are needed to produce 9.5 moles of Fe(s)?

C) Which of the following equations correctly describes the combustion of CH₄and O₂to produce water (H₂O) and carbon dioxide (CO₂)?

a)     CH₄ + O₂  CO₂ + H₂O

b)     CH₄ + O₂  CO₂ + 2H₂O

c)     CH₄ + 2O₂  CO₂ + 2H₂O

d)     CH₄ + 3O₂  2CO₂ + H₂O

e)     2CH₄ + 3O₂  2CO₂ + 2H₂O

D) What is the coefficient for oxygen when the following equation is balanced?

NH₃(g) + O₂(g) → NO₂(g) + H₂O(g)

Answer 1

A)
2H2 + O2 ---> 2H2O
number of moles of O2 present = mass / molar mass
                        = 30.8 / 32
                        = 0.9625 mol

number of moles of H2 present = mass / molar mass
                        = 30.8 / 2
                        = 15.4 mol

Clearly O2 is limiting reagent. I will use O2 in my further calculations.
1 mol of O2 gives 2 mol fo H2O.

So, number of moles of H2O = 2* moles of O2
   = 2* 0.9625
   = 1.925 mol
mass of water = mol * molar mass
    = 1.925*18
    = 34.65 gm

B)
2 mol of Fe requires 3 mol of CO
9.5 mol of Fe will require = 3*9.5/2 = 14.25 mol of CO
2 mol of CO required = 1 mol O2
so, number of moles of O2 required = 14.25 / 2 = 7.125 mol

c)
Answer: c

D)
4 NH3 + 7 O2 --> 4 NO2 +6 H2O
Answer: 7