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Chapter 5: Problem 63

(a) What volumes (in liters) of ammonia and oxygen must react to form \(12.8 \mathrm{~L}\) of nitric oxide according to the equation at the same temperature and pressure? $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ (b) What volumes (in liters) of propane and water vapor must react to form \(8.96 \mathrm{~L}\) of hydrogen according to the equation at the same temperature and pressure? $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{CO}(g)+7 \mathrm{H}_{2}(g) $$

Short Answer

Expert verified
For part (a), 12.8 L of \(NH_{3}\) and 16 L of \(O_{2}\) are required. For part (b), 1.254 L of \(C_{3}H_{8}\) and 3.853 L of \(H_{2}O\) are required.

Step by step solution

01

Establish Mole Ratios for the First Reaction

From the balanced chemical equation \(4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)\), we can establish that 4 volumes of \(NH_{3}\), and 5 volumes of \(O_{2}\), react to form 4 volumes of \(NO\). This means ratio of the volume of \(NH_{3}\) to the volume of \(NO\) is 1 (i.e., 4/4), and the ratio of the volume of \(O_{2}\) to the volume of \(NO\) is 1.25 (i.e., 5/4).
02

Calculate Volumes for the First Reaction

We know \(12.8 L\) of \(NO\) are formed. So using the volume ratios established in step 1, the volume of \(NH_{3}\) required would be \(12.8 L * 1 = 12.8 L\), and the volume of \(O_{2}\) required would be \(12.8 L * 1.25 = 16 L\).
03

Establish Mole Ratios for the Second Reaction

From the balanced chemical equation \(C_{3}H_{8}(g) + 3H_{2}O(g) \rightarrow 3CO(g) + 7H_{2}(g)\), we can establish that 1 volume of \(C_{3}H_{8}\), and 3 volumes of \(H_{2}O\), react to form 7 volumes of \(H_{2}\). This means the ratio of the volume of \(C_{3}H_{8}\) to the volume of \(H_{2}\) is 0.14 (i.e., 1/7), and the ratio of the volume of \(H_{2}O\) to the volume of \(H_{2}\) is 0.43 (i.e., 3/7).
04

Calculate Volumes for the Second Reaction

We know \(8.96 L\) of \(H_{2}\) are formed. So using the volume ratios established in step 3, the volume of \(C_{3}H_{8}\) required would be \(8.96 L * 0.14 = 1.254 L\), and the volume of \(H_{2}O\) required would be \(8.96 L * 0.43 = 3.853 L\).

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Most popular questions from this chapter

Under the same conditions of temperature and pressure, which of the following gases would behave most ideally: Ne, \(\mathrm{N}_{2}\), or \(\mathrm{CH}_{4}\) ? Explain.

Cite two pieces of evidence to show that gases do not behave ideally under all conditions.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) burns in air: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Balance the equation and determine the volume of air in liters at \(35.0^{\circ} \mathrm{C}\) and \(790 \mathrm{mmHg}\) required to burn \(227 \mathrm{~g}\) of ethanol. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume.

The following procedure is a simple though somewhat crude way to measure the molar mass of a gas. A liquid of mass \(0.0184 \mathrm{~g}\) is introduced into a syringe like the one shown here by injection through the rubber tip using a hypodermic needle. The syringe is then transferred to a temperature bath heated to \(45^{\circ} \mathrm{C},\) and the liquid vaporizes. The final volume of the vapor (measured by the outward movement of the plunger) is \(5.58 \mathrm{~mL}\) and the atmospheric pressure is \(760 \mathrm{mmHg}\). Given that the compound's empirical formula is \(\mathrm{CH}_{2}\), determine the molar mass of the compound.

Commercially, compressed oxygen is sold in metal cylinders. If a 120-L cylinder is filled with oxygen to a pressure of 132 atm at \(22^{\circ} \mathrm{C},\) what is the mass (in grams) of \(\mathrm{O}_{2}\) present? How many liters of \(\mathrm{O}_{2}\) gas at 1.00 atm and \(22^{\circ} \mathrm{C}\) could the cylinder produce? (Assume ideal behavior.)
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