Question

O2 + 2C0-2CO2

Answer 1

Answer: 1. $CO$ is the limiting reagent and $O_2$ is the excess reagent.

2. 2.91 g of excess reagent

3. 4.66 g of carbon dioxide produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} CO=\frac{2.98g}{28g/mol}=0.106moles$

$\text{Moles of} O_2=\frac{4.62g}{32g/mol}=0.144moles$

$O_2+2CO\rightarrow 2CO_2$  

According to stoichiometry :

2 moles of $CO$ require = 1 mole of $O_2$

Thus 0.106 moles of $CO$ will require=$\frac{1}{2}\times 0.106=0.053moles$  of $O_2$

Thus $CO$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles

Mass of excess reagent $O_2=moles\times {\text {Molar mass}}=0.091moles\times 32g/mol=2.91g$

As 2 moles of $CO$ give = 2 moles of $CO_2$

Thus 0.106 moles of $CO$ give =$\frac{2}{2}\times 0.106=0.106moles$  of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=0.106moles\times 44g/mol=4.66g$