in the combustion of carbon monoxide, what mass of co is required to produce 0.39 grams of carbon dioxide…

in the combustion of carbon monoxide, what mass of co is required to produce 0.39 grams of carbon dioxide? the unbalanced equation is shown below: co (g) + o₂ (g) → co₂ (g)\nco 28.01 g/mol\no₂ 32.0 g/mol\nco₂ 44.0 g/mol\n0.39g\n0.25g\n0.61g\n0.75g

in the combustion of carbon monoxide, what mass of co is required to produce 0.39 grams of carbon dioxide? the unbalanced equation is shown below: co (g) + o₂ (g) → co₂ (g)\nco 28.01 g/mol\no₂ 32.0 g/mol\nco₂ 44.0 g/mol\n0.39g\n0.25g\n0.61g\n0.75g

Answer

Explanation:

Step1: Balance the chemical equation

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

Step2: Calculate moles of $CO_2$

The molar - mass of $CO_2$ is $M_{CO_2}=44.0\ g/mol$. The number of moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}$. Given $m_{CO_2} = 0.39\ g$, so $n_{CO_2}=\frac{0.39\ g}{44.0\ g/mol}\approx0.00886\ mol$.

Step3: Determine moles of $CO$ from the stoichiometry

From the balanced equation, the mole - ratio of $CO$ to $CO_2$ is $1:1$. So, $n_{CO}=n_{CO_2}= 0.00886\ mol$.

Step4: Calculate mass of $CO$

The molar - mass of $CO$ is $M_{CO}=28.01\ g/mol$. The mass of $CO$, $m_{CO}=n_{CO}\times M_{CO}=0.00886\ mol\times28.01\ g/mol\approx0.25\ g$.

Answer:

0.25g