Iron(III) oxide reacts with carbon to give iron and carbon monoxide.

$F{e}_2{O}_3\left(s\right)+3C\left(s\right)\to 2Fe\left(s\right)+3CO\left(g\right)$

a. How many grams of $C$are required to react with localid="1653488683706" $16.5\mathrm{g}$of $F{e}_2{O}_3?$

b. How many grams of $CO$are produced whenlocalid="1653488713236" $36.0\mathrm{g}$of $C$reacts ?

c. How many grams of Fe can be produced when localid="1653488730008" $6.00\mathrm{g}$of $F{e}_2{O}_3$reacts?

We have to find out the how many grams of $C$required to react with localid="1653488895828" $16.5\mathrm{g}$of $F{e}_2{O}_3$.

Now, we find the solution of this questions,

$165\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3$reacts with $36\mathrm{g}\mathrm{C}$,

$1\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3$reacts with $\frac{36\cancel{\mathrm{g}}}{165\cancel{\mathrm{g}}}\mathrm{C}$,

So, $16.5\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3$reacts with $\frac{36\cancel{\mathrm{g}}\times 16.5\mathrm{g}}{165\cancel{\mathrm{g}}}\mathrm{C}$,

$=3.6\mathrm{g}\mathrm{C}$

We have to find out the how many grams of$CO$are produced whenlocalid="1653489119062" $36.0\mathrm{g}$of$C$reacts.

Now, we solve this question,

$F{e}_2{O}_3\left(s\right)+3C\left(s\right)\to 2Fe\left(s\right)+3CO\left(g\right)$,

localid="1653489261891" role="math" $1\mathrm{mol}\mathrm{C}=12\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$,

localid="1653489276753" role="math" $3\mathrm{mol}\mathrm{C}=12\frac{\mathrm{g}}{\cancel{\mathrm{mol}}}\times 3\cancel{\mathrm{mol}}=36\mathrm{g}$,

So,localid="1653489291299" $36\mathrm{g}\mathrm{C}$producedlocalid="1653489317581" $84\mathrm{g}\mathrm{CO}$.

We have to find out the how many grams of$Fe$can be produced whenlocalid="1653489352411" $6.00\mathrm{g}$of ${\mathrm{Fe}}_2{\mathrm{O}}_3$reacts.

We know that

$1\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3$produces $\frac{175\cancel{\mathrm{g}}}{165\cancel{\mathrm{g}}}\mathrm{Fe}$,

$\therefore 6\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3$produces $\frac{117\cancel{\mathrm{g}}\times 6\mathrm{g}}{165\cancel{\mathrm{g}}}\mathrm{Fe}$,

$=4.254\mathrm{g}\mathrm{Fe}$.