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Question 19
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
Step 1
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:
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:
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:
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:
The molar mass of N2 is approximately 28 g/mol:
Thus, the products of the reaction are:
The correct answer is therefore D: 8.8 g of CO2 and 2.8 g of N2.
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