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Chapter 1: Problem 8

Fill in the blanks. The ratio of the volumes of \(11 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28 \mathrm{~g}\) of \(\mathrm{CO}\) at \(\mathrm{STP}\) is

Short Answer

Expert verified
Answer: The ratio of the volumes of 11 g of CO₂ and 28 g of CO at STP is 7:1.

Step by step solution

01

Calculate the number of moles of each gas

First, we need to find the number of moles of each gas. We can do this by dividing their given masses by their molar masses. The molar mass of \(\mathrm{CO}_{2}\) is 44 g/mol and the molar mass of \(\mathrm{CO}\) is 28 g/mol. Number of moles of \(\mathrm{CO}_{2}\) = \(\frac{11\mathrm{~g}}{44\mathrm{~g/mol}}\) Number of moles of \(\mathrm{CO}\) = \(\frac{28\mathrm{~g}}{28\mathrm{~g/mol}}\)
02

Calculate the volumes of each gas

Now, we will calculate the volumes of each gas using the number of moles and the molar volume of an ideal gas at STP (22.4 liters/mol). Volume of \(\mathrm{CO}_{2}\) = Number of moles of \(\mathrm{CO}_{2}\) × Molar volume at STP Volume of \(\mathrm{CO}_{2}\) = \(\frac{11\mathrm{~g}}{44\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}\) Volume of \(\mathrm{CO}\) = Number of moles of \(\mathrm{CO}\) × Molar volume at STP Volume of \(\mathrm{CO}\) = \(\frac{28\mathrm{~g}}{28\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}\)
03

Find the ratio of the volumes

Finally, we will find the ratio of the volumes of the two gases. We can do this by dividing the volume of \(\mathrm{CO}_{2}\) by the volume of \(\mathrm{CO}\). Ratio = \(\frac{\text{Volume of }\mathrm{CO}_{2}}{\text{Volume of }\mathrm{CO}}\) Ratio = \(\frac{\frac{11\mathrm{~g}}{44\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}}{\frac{28\mathrm{~g}}{28\mathrm{~g/mol}} \times 22.4\mathrm{~L/mol}}\) Ratio = \(\frac{11}{44} \times \frac{28}{1}=\boxed{7}\) So, the ratio of the volumes of \(11 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28 \mathrm{~g}\) of \(\mathrm{CO}\) at STP is 7:1.

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