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Chapter 6: Problem 143

Carbon monoxide, \(\mathrm{CO}\), and hydrogen react according to the equation below. $$3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of which reactant gas remains if \(12.0 \mathrm{LCO}(\mathrm{g})\) and \(25.0 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) are allowed to react? Assume that the volumes of both gases are measured at the same temperature and pressure.

Short Answer

Expert verified
The limiting reactant is H2 and after the reaction, 1.3 L of CO remains.

Step by step solution

01

Identify the Mole Ratio

The balanced equation tells us that 3 moles of CO react with 7 moles of H2 to give products. So, the mole ratio of CO to H2 is 3:7.
02

Calculate Reactant Ratios

According to the given exercise, we have 12.0 L of CO and 25.0 L of H2. Considering the mole ratio from step 1, we can say that for every 3 L of CO we need 7 L of H2.
03

Determine the Limiting Reactant

To find the limiting reactant, compare the ratio of reactants provided with the ratio required by the equation. In this case, (25.0L H2)/(12.0L CO) = 2.08 and (7 moles H2)/(3 moles CO) = 2.33. Since the given ratio is less than the required ratio, we know that H2 is the limiting reactant, and CO is in excess.
04

Calculate the Remaining Volume of CO

Since we know that H2 is the limiting reactant and 25.0 L of H2 react completely, according to the ratio in the balanced equation, we would need (3 moles CO/ 7 moles H2) x 25.0 L H2 = 10.7 L CO. Initially, we had 12.0 L of CO, so the remaining CO would be 12.0 L - 10.7 L = 1.3 L.

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Most popular questions from this chapter

The gas with the greatest density at STP is (a) \(\mathrm{N}_{2} \mathrm{O}\) (b) \(\mathrm{Kr} ;\) (c) \(\mathrm{SO}_{3} ;\) (d) \(\mathrm{Cl}_{2}\).

Aluminum (Al) and iron (Fe) each react with hydrochloric acid solution (HCl) to produce a chloride salt and hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}) .\) A \(0.1924 \mathrm{g}\) sample of a mixture of \(\mathrm{Al}\) and \(\mathrm{Fe}\) is treated with excess \(\mathrm{HCl}\) solution. A volume of \(159 \mathrm{mL}\) of \(\mathrm{H}_{2}\) gas is collected over water at \(19.0^{\circ} \mathrm{C}\) and 841 Torr. What is the percent (by mass) of Fe in the mixture? The vapor pressure of water at \(19.0^{\circ} \mathrm{C}\) is 16.5 Torr.

A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]

According to the CRC Handbook of Chemistry and Physics (83rd ed.), the molar volume of \(\mathrm{O}_{2}(\mathrm{g})\) is \(0.2168 \mathrm{Lmol}^{-1}\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). (Note: \(1 \mathrm{MPa}=\) \(\left.1 \times 10^{6} \mathrm{Pa} .\right)\)(a) Use the van der Waals equation to calculate the pressure of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) if the volume is 0.2168 L. What is the \% error in the calculated pressure? The van der Waals constants are \(a=1.382 \mathrm{L}^{2}\) bar \(\mathrm{mol}^{-2}\) and \(b=0.0319 \mathrm{L} \mathrm{mol}^{-1}\) (b) Use the ideal gas equation to calculate the volume of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). What is the \% error in the calculated volume?

A 10.0 g sample of a gas has a volume of \(5.25 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(762 \mathrm{mm} \mathrm{Hg} .\) If \(2.5 \mathrm{g}\) of the same gas is added to this constant 5.25 L volume and the temperature raised to \(62^{\circ} \mathrm{C},\) what is the new gas pressure?
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