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netineya [11]
3 years ago
10

Problem PageQuestion Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume

of carbon tetrachloride would be produced by this reaction if of chlorine were consumed?
Chemistry
1 answer:
IgorC [24]3 years ago
8 0

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.

Be sure your answer has the correct number of significant digits.

Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm  (at STP)

V = Volume of gas = 3.16 L

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K=

n=\frac{PV}{RT}

n=\frac{1atm\times 3.16L}{0.0820 L atm/K mol\times 273K}=0.141moles

CH_4+4Cl_2\rightarrow 4HCl+CCl_4

According to stoichiometry:

4 moles of chlorine produces = 1 mole of carbon tetrachloride

Thus 0.141 moles of methane produces = \frac{1}{4}\times 0.141=0.0352 moles of carbon tetrachloride

volume of carbon tetrachloride =moles\times {\text {Molar volume}}=0.0352mol\times 22.4L/mol=0.788L

Thus volume of carbon tetrachloride that would be produced is 0.788 L

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Approximately 1.854\; \rm mol\cdot L^{-1}.

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<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of \rm Sr, \rm O, and \rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

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Calculate the formula mass of \rm Sr(OH)_2:

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<h3>Number of moles of strontium hydroxide in the solution</h3>

M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1} means that each mole of \rm Sr(OH)_2 formula units have a mass of 121.634\; \rm g.

The question states that there are 10.60\; \rm g of \rm Sr(OH)_2 in this solution.

How many moles of \rm Sr(OH)_2 formula units would that be?

\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}.

<h3>Molarity of this strontium hydroxide solution</h3>

There are 8.71467\times 10^{-2}\; \rm mol of \rm Sr(OH)_2 formula units in this 47\; \rm mL solution. Convert the unit of volume to liter:

V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:

\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}.

(Rounded to four significant figures.)

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