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.
