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Chapter 4: Problem 80

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

Short Answer

Expert verified
The mass of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) formed in reaction (b) is estimated to be around \(501.67\) g.

Step by step solution

01

Calculate the number of moles of CH4

The initial mass of \(\mathrm{CH}_{4}\) is given as 112 g. The molar mass of \(\mathrm{CH}_{4}\) is approximately 16.04 g/mol. Therefore, the initial amount of \(\mathrm{CH}_{4}\) in moles is \(moles = \frac{mass}{molar\,mass}\) = \(\frac{112\, g}{16.04 \,g/mol} = 6.98 \,moles\)
02

Calculate the number of moles of CH3Cl produced in reaction (a)

The yield of reaction (a) is 92%, thus the actual amount of \(\mathrm{CH}_{3}\mathrm{Cl}\) produced is \((6.98 moles)*0.92 = 6.42 moles\)
03

Calculate the number of moles of CH2Cl2 produced in reaction (b)

The yield of reaction (b) is also 92%, thus the actual amount of CH2Cl2 that is produced is \((6.42 moles)*0.92 = 5.91 moles\)
04

Convert the number of moles of CH2Cl2 to mass

The molar mass of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) is approximately 84.93 g/mol. Therefore, the final amount of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) in grams is \(grams = moles * molar\,mass = (5.91\, moles) * (84.93 \,g/mol) = 501.67 g\)

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Most popular questions from this chapter

A commercial method of manufacturing hydrogen involves the reaction of iron and steam. $$ 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) $$ (a) How many grams of \(\mathrm{H}_{2}\) can be produced from \(42.7 \mathrm{g}\) Fe and an excess of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (steam)? (b) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are consumed in the conversion of \(63.5 \mathrm{g}\) Fe to \(\mathrm{Fe}_{3} \mathrm{O}_{4} ?\) (c) If \(14.8 \mathrm{g} \mathrm{H}_{2}\) is produced, how many grams of \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) must also be produced?

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