5. if 300 g of fe₂o₃ and 84 g of co react as described by the equation fe₂o₃(s) + 3co(g) → 2fe(l) + 3co₂(g)…

Answer:

CO

Explanation:

Step1: Calculate molar masses

$M_{Fe_2O_3}=(2\times55.85 + 3\times16)=159.7\ g/mol$, $M_{CO}=12 + 16 = 28\ g/mol$

Step2: Calculate moles of reactants

$n_{Fe_2O_3}=\frac{300\ g}{159.7\ g/mol}\approx1.88\ mol$, $n_{CO}=\frac{84\ g}{28\ g/mol}=3\ mol$

Step3: Determine mole - ratio from the equation

The mole - ratio of $Fe_2O_3$ to $CO$ in the reaction is $1:3$.

Step4: Find the required moles of CO for given $Fe_2O_3$

For $1.88\ mol$ of $Fe_2O_3$, the required moles of $CO$ is $n_{CO - required}=1.88\ mol\times3 = 5.64\ mol$

Step5: Compare available and required moles of CO

We have only $3\ mol$ of $CO$ available, which is less than the required $5.64\ mol$. So, CO is the limiting reactant.