##### Stoichiometry Part II: Tackling interesting A Level Chemistry Prelim Question: Gases

- focuschemistry
- August 28, 2017
- Stoichiometry and Mole Concept

In this section, we shall look at a few past year MCQ questions from top JCs.

In the A levels Chemistry, the MCQ paper (paper 1) is a 40-question paper, in which students are given only 60 minutes to work out the solutions. That works out to 1.5 min / question.

A common problem amongst A level students is trying to complete the paper on time.

Hence, the challenge in paper 1 is not only to solve the question correctly, but to do so in a very quick manner.

Here we present 2 ways to do this question:

**RI 2006, Paper 1:**

Methane was burned in an incorrectly adjusted burner. The methane was converted into a mixture of carbon dioxide and carbon monoxide in the ratio of 98:2, together with water vapour.

What will be the volume of oxygen consumed when y dm^{3} of methane is burned?

| B. (2y-0.01y)dm^{3} | |||||

| D. (y-0.01y)dm^{3} |

**Solutions:**

There are 2 ways to solve this question. One of the methods is presented below:

__Method 1:__

Balance the overall equation for the combustion of CH_{4} to produce CO_{2} and CO in 98:2 ratio:

100CH_{4} + 199O_{2} → __98__CO_{2} + __2__CO + 200H_{2}O

Reacted : y dm^{3}

Volume of O_{2} reacted = |
| x199 = 1.99 y dm^{3} |

Answer is **B** (2y – 0.01y = 1.99 y)

__Method 2:__

Break up the overall equation into 2 equations:

CH_{4} + O_{2} → CO_{2} + H_{2}O ………(1)

CH_{4} + O_{2} → CO + H_{2}O ………(2)

Since CO_{2}:CO = 98:2,

0.98y dm^{3} of CH_{4} undergoes reaction (1) and 0.02 y dm^{3} CH_{4} undergoes reaction (2)

The total volume of O_{2} needed can then be calculated and the answer is still **B**.

**RI, 2004, JC1BT1**

A 10 cm^{3} sample of a gaseous hydrocarbon measured at room temperature and pressure (r.t.p) was reacted with 70 cm^{3} of oxygen.

The resulting gas mixture collected was found to have a volume of 60 cm^{3} (also measured at r.t.p). The gas mixture was completely absorbed by aqueous potassium hydroxide.

Calculate the empirical formula of the hydrocarbon

A. C_{2}H_{3} | B. C_{3}H_{2} | C. C_{4}H_{6} | D. C_{6}H_{4} |

**Solutions**

Let the formula of the unknown hydrocarbon be C_{x}H_{y}

The gas absorbed by potassium hydroxide is CO_{2}.

Equation:

C_{x}H_{y} + ( |
| )O_{2} → xCO_{2} + |
| H_{2}O |

Or: | C_{x}H_{y} + ( |
| + x)O_{2} → xCO_{2} + |
| H_{2}O |

Reacted (cm^{3}): | 10 | 70 | 60 |

Mole ratio of | CH_{4} | : | CO_{2} | |

1 | : | X | ||

10 | : | 60 |

X = 6

Mole ratio of | CH_{4} | : | CO_{2} | |||||

1 | : |
| ||||||

10 | : | 70 |

⇒ |
| + X = 7 |

⇒ y = 4(7-6) = 4

Thus the empirical formula of the hydrocarbon is C_{6}H_{4}. Answer is D.

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