Answer to Question #141158 in General Chemistry for mike

Question

Calculate the average rate of production of carbon dioxide in mol dm-3 s-1 if 0.480g of carbon dioxide is produced in a 2.00dm3 flask in 1.06 minutes, according to the reaction:

2CO(g) + O2(g) → 2CO2(g)

Answer 1

$\begin{aligned} \textsf{mass of }2CO_2 \textsf{ produced } &= 0.480g\\ \textsf{mass of }CO_2 \textsf{ produced } &= \dfrac{\textsf{mass of }2CO_2 \textsf{ produced }}{2}\\ \textsf{mass of }CO_2 \textsf{ produced } &= 0.240g\\ \\ \textsf{molar mass of }CO_2 &= 44g/mol\\ \\ \textsf{number of moles (n)} &= \dfrac{mass}{molar \ mass}\\ \\ &= \dfrac{0.240}{44}\\ \\ &= 0.00545\, mol \end{aligned}$

$\textsf{volume of container }(V) = 2.00dm^3$

$\begin{aligned} \textsf{Molar Concentration (C) } &= \dfrac{n}{V}\\ \\ &= \dfrac{0.00545}{2}\\ \\ &= 0.002725\, moldm^{-3} \end{aligned}$

$\begin{aligned} \textsf{time taken (T) } &=1.06\ minutes\\ &= 1.06(60\ seconds)\\ &= 63.6\ seconds \end{aligned}$

$\begin{aligned} \textsf{rate of production } &= \dfrac{C}{T}\\ \\ &= \dfrac{0.002725\ moldm^{-3}}{63.6\ s}\\ \\ &= 4.285 × 10^{-5}\ mol/dm^3s \end{aligned}$