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Chapter 9: Problem 60

Consider the unbalanced chemical equation \(\mathrm{NO}+\mathrm{O}_{2} \rightarrow \mathrm{NO}_{2}\) (a) Balance the equation. (b) Translate the equation into words using the word mole(s) wherever you can. (c) To produce 2 moles of \(\mathrm{NO}_{2}\) by the reaction you just wrote, how many grams of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) must you combine? (d) What is the theoretical yield in grams of \(\mathrm{NO}_{2}\) ? (e) You carry out the reaction and recover \(22.5 \mathrm{~g}\) of \(\mathrm{NO}_{2}\). What is the percent yield?

Short Answer

Expert verified
The balanced chemical equation for the reaction is \(2 NO + O_2 \rightarrow 2 NO_2\). To produce 2 moles of NO2, 60.02 g of NO and 32.00 g of O2 are needed. The theoretical yield of NO2 is 92.02 g, and the percent yield of the reaction is 24.45%.

Step by step solution

01

1. Balance the chemical equation

To balance the given equation, \(NO + O_2 \rightarrow NO_2\), we need to make sure that the number of atoms for each element on both sides of the equation is equal. The balanced equation is: \(2 NO + O_2 \rightarrow 2 NO_2\)
02

2. Translate the equation into words

The balanced equation can be translated into words as: "2 moles of nitrogen monoxide react with 1 mole of oxygen gas to form 2 moles of nitrogen dioxide."
03

3. Calculate grams of NO and O2

To produce 2 moles of NO2, we need 2 moles of NO and 1 mole of O2. We will now find the mass of each reactant. For NO: Molar mass of NO = (1 x N's atomic mass) + (1 x O's atomic mass) = 14.01 + 16.00 = 30.01 g/mol Mass of 2 moles of NO = (2 mol) x (30.01 g/mol) = 60.02 g For O2: Molar mass of O2 = 2 x O's atomic mass = 2 x 16.00 = 32.00 g/mol Mass of 1 mole of O2 = (1 mol) x (32.00 g/mol) = 32.00 g So, we need 60.02 g of NO and 32.00 g of O2 to produce 2 moles of NO2.
04

4. Theoretical yield in grams of NO2

To find the theoretical yield in grams of NO2, we need to calculate the mass of 2 moles of NO2. Molar mass of NO2 = (1 x N's atomic mass) + (2 x O's atomic mass) = 14.01 + 2 x 16.00 = 46.01 g/mol Mass of 2 moles of NO2 = (2 mol) x (46.01 g/mol) = 92.02 g The theoretical yield of NO2 is 92.02 g.
05

5. Calculate the percent yield

The actual yield is given as 22.5 g of NO2. To calculate the percent yield, we can use the formula: Percent yield = (actual yield / theoretical yield) x 100 Percent yield = (22.5 g / 92.02 g) x 100 = 24.45% The percent yield of the reaction is 24.45%.

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