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In a catalytic converter 5.6 g of carbon monoxide reacts with an excess of nitrogen monoxide - CIE - A-Level Chemistry - Question 19 - 2019 - Paper 1

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In a catalytic converter 5.6 g of carbon monoxide reacts with an excess of nitrogen monoxide. What is produced in this reaction? A 2.4 g of C and 6.0 g of NO2 B 2.... show full transcript

Worked Solution & Example Answer:In a catalytic converter 5.6 g of carbon monoxide reacts with an excess of nitrogen monoxide - CIE - A-Level Chemistry - Question 19 - 2019 - Paper 1

Step 1

What is produced in this reaction?

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Answer

To determine the products of this reaction, we can start by writing the balanced chemical equation for the reaction of carbon monoxide (CO) with nitrogen monoxide (NO).

The reaction can be represented as:

2extCO+extNO→2extCO2+N22 ext{CO} + ext{NO} \rightarrow 2 ext{CO}_2 + \text{N}_2

Given that we have 5.6 g of CO, we need to calculate how many moles of CO this represents. The molar mass of CO is approximately 28 g/mol:

Moles of CO=5.6 g28 g/mol=0.2 moles\text{Moles of CO} = \frac{5.6 \text{ g}}{28 \text{ g/mol}} = 0.2 \text{ moles}

Using the balanced equation, 2 moles of CO yield 2 moles of CO2. Therefore, 0.2 moles of CO will produce 0.2 moles of CO2.

Now we calculate the mass of CO2 produced:

Mass of CO2=0.2 moles×44 g/mol=8.8 g\text{Mass of CO}_2 = 0.2 \text{ moles} \times 44 \text{ g/mol} = 8.8 \text{ g}

Since we react with an excess of NO, we also produce nitrogen gas (N2) in a stoichiometric ratio. The balanced equation shows that for every 2 moles of CO, 1 mole of N2 is produced. Hence, 0.2 moles of CO will yield:

Moles of N2=0.22=0.1 moles\text{Moles of N}_2 = \frac{0.2}{2} = 0.1 \text{ moles}

The molar mass of N2 is approximately 28 g/mol:

Mass of N2=0.1 moles×28 g/mol=2.8 g\text{Mass of N}_2 = 0.1 \text{ moles} \times 28 \text{ g/mol} = 2.8 \text{ g}

Thus, the products of the reaction are:

  • 8.8 g of CO2
  • 2.8 g of N2

The correct answer is therefore D: 8.8 g of CO2 and 2.8 g of N2.

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