Let's start by writing the equation for this reaction:
"X" + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)
Using the fact that 0.112 "dm"^3 of compound "X" was used up, we can use the ideal-gas equation to solve for the number of moles of "X":
PV = nRT
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P = 1 "bar" = 0.987 "atm" (standard pressure)
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V = 0.112 "dm"^3 = 0.112 "L" (given)
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R = 0.082057("L"·"atm")/("mol"·"K") (gas constant)
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T = 273.15 "K" (standard temperature)
Plugging in these values and solving for n, we have
n = (PV)/(RT) =((0.987cancel("atm"))(0.112cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(273.15cancel("K"))) = color(red)(0.00493 color(red)("mol X"
What we can do now is find the number of moles of "CO"_2 present, using the given mass and its molar mass:
0.88cancel("g CO"_2)((1color(white)(l)"mol CO"_2)/(44.01cancel("g CO"_2))) = color(green)(0.020 color(green)("mol CO"_2
The number of carbon atoms in "X" is equal to the ratio of moles of "CO"_2 to moles of "X" (the coefficient in front of "CO"_2 is equal to the number of "C" atoms in "X"). Therefore, the number of "C" atoms in compound "X" is
"atoms C" = ("mol CO"_2)/("mol X") = (color(green)(0.020"mol CO"_2))/(color(red)(0.00493"mol X")) = 4.1 ~~ color(blue)(4
Thus, there must be 4 atoms of "C" in compound "X", so option (d) is correct.
