# Question a418b

Jun 19, 2017

(d) $4$ $\text{atoms C}$

#### Explanation:

Let's start by writing the equation for this reaction:

$\text{X" + "O"_2(g) rarr "CO"_2(g) + "H"_2"O} \left(g\right)$

Using the fact that $0.112$ ${\text{dm}}^{3}$ of compound $\text{X}$ was used up, we can use the ideal-gas equation to solve for the number of moles of $\text{X}$:

$P V = n R T$

• $P = 1$ $\text{bar} = 0.987$ $\text{atm}$ (standard pressure)

• $V = 0.112$ ${\text{dm}}^{3} = 0.112$ $\text{L}$ (given)

• $R = 0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$ (gas constant)

• $T = 273.15$ $\text{K}$ (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 ${\text{CO}}_{2}$ 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 $\text{X}$ is equal to the ratio of moles of ${\text{CO}}_{2}$ to moles of $\text{X}$ (the coefficient in front of ${\text{CO}}_{2}$ is equal to the number of $\text{C}$ atoms in $\text{X}$). Therefore, the number of $\text{C}$ atoms in compound $\text{X}$ 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 $\text{C}$ in compound $\text{X}$, so option (d) is correct.