Question #201146

When CO2(g)CO2(g) is put in a sealed container at 730 KK and a pressure of 10.0 atm and is heated to 1420 KK , the pressure rises to 24.1 atm. Some of the CO2CO2 decomposes to COCO and O2O2.

Calculate the mole percent of CO2 that decomposes.

**Express your answer using two significant figures.**

%CO2 =

Expert's answer

Ideal Gas Law:

pV=nRT

R = 0.08206 L×atm/mol×K

"10.0 \\times V = n \\times 0.08206 \\times 730 \\\\\n\n10.0 \\times V = 59.90n \\\\\n\nn = 0.1669V"

After decomposition:

"24.1 \\times V = n^* \\times 0.08206 \\times 1420 \\\\\n\n24.1 \\times V = 116.52 \\times n^* \\\\\n\nn^* = 0.2068V"

(which is the total amount of moles of the gases)

Let's suppose, V = 1L, so:

"n=0.1669 \\\\\n\nn^*=0.2068"

For the decomposition reaction, we can do a reaction table:

2CO_{2} → 2CO + O_{2}

0.1669____ 0___ 0

-2x______ +2x__ +x

0.1669-2x __2x__x

"n^* = 0.1669 -2x + 2x +x \\\\\n\n0.1669 + x = 0.2068 \\\\\n\nx = 0.0399"

So, "2 \\times 0.0399 = 0.0798" mol of CO_{2} decomposes, the percent is:

"\\frac{0.0798}{0.1669} \\times 100 \\;\\% = 47.81 \\; \\%"

Answer: 47.81 %

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