Respuesta :
Answer: The maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is [tex]O_2[/tex] and the mass of excess reagent (carbon) remained is 4.644 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For carbon:
Given mass of carbon = 9.37 g
Molar mass of carbon = 12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon}=\frac{9.37g}{12g/mol}=0.781mol[/tex]
- For oxygen gas:
Given mass of oxygen gas = 12.6 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{12.6g}{32g/mol}=0.394mol[/tex]
The chemical equation for the reaction of carbon and oxygen gas follows:
[tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]
By Stoichiometry of the reaction:
1 moles of oxygen gas reacts with 1 mole of carbon metal.
So, 0.394 moles of oxygen gas will react with = [tex]\frac{1}{1}\times 0.394=0.394mol[/tex] of carbon metal.
As, given amount of carbon metal is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (carbon) left = [0.781 - 0.394] = 0.387 moles
By Stoichiometry of the reaction:
1 moles of oxygen gas produces 1 mole of carbon dioxide
So, 0.394 moles of oxygen gas will produce = [tex]\frac{1}{1}\times 0.394=0.394moles[/tex] of carbon dioxide
Now, calculating the mass of carbon and carbon dioxide from equation 1, we get:
- For carbon:
Excess moles of carbon = 0.387 moles
Molar mass of carbon = 12 g/mol
Putting values in equation 1, we get:
[tex]0.387mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=(0.387mol\times 12g/mol)=4.644g[/tex]
- For carbon dioxide:
Moles of carbon dioxide = 0.394 moles
Molar mass of carbon dioxide = 44 g/mol
Putting values in equation 1, we get:
[tex]0.394mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.394mol\times 4g/mol)=17.34g[/tex]
Hence, the maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is [tex]O_2[/tex] and the mass of excess reagent (carbon) remained is 4.644 grams
