Question

Consider a reaction.Fe2O3+3CO=2Fe+3CO2. 10 g of Fe2O2 is reacted with 9 g of CO2. Find the limiting reactant. How many moles of unreacted reactant is left over? Calculate the mole of CO consumed in the reaction.what mass of NaOH is required to absorb whole CO2 produced?

Answer 1

Answer:

please check those two pictures for the answer.

Answer 2

Concept Introduction:-

It may be in the form of a word, a symbol, or a figure that reflects the arithmetic value of a quantity.

Given Information:-

We have been given that $Fe_2O_3+3CO=2Fe+3CO_2$. $10$ g of $Fe_2O_2$ is reacted with $9$ g of $CO_2$.

To Find:-

We have to find that the limiting reactant. How many moles of unreacted reactant is left over? Calculate the mole of CO consumed in the reaction.what mass of NaOH is required to absorb whole $CO_2$ produced?

Solution:-

According to the problem

For starters, a very quick way of figuring out which reactant will act as the limiting reagent when dealing with equal masses of the two reactants is to look at their molar masses.

When you're dealing with equal masses of two reactants, the reactant that has the greater molar mass will have the smallest number of moles present in the given mass.

In this case, iron (III) oxide has a molar mass of

$159.7$ g mol$^{-1}$ and carbon monoxide has a molar mass of $28.01$ g mol$^{-1}$

Right from the start, you should be able to tell that iron(III) oxide will be the limiting reagent because you have significantly more moles of carbon monoxide in $2.00$ kg than moles of iron(III) oxide in $2.00$ kg.

More specifically, you will have

$2000 g \cdot \frac{1 \text { mole } \mathrm{Fe}_{2} \mathrm{O}_{3}}{159.7 g}=12.52$ moles $\mathrm{Fe}_{2} \mathrm{O}_{3}$ $2000 g \cdot \frac{1 \text { mole CO }}{28.01 g}=71.40$ moles $\mathrm{CO}$

If all the moles of iron(III) were to take part in the reaction, then the reaction would also consume $12.52$ moles $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \frac{3 \text { moles } \mathrm{CO}}{1 \text { mole- } \mathrm{Fe}_{2} \mathrm{O}_{3}}=37.56$ moles $\mathrm{CO}$

The rest of the moles of carbon monoxide will be in excess, i.e. they will not take part in the reaction.

For the last part, you got the approach down correctly. All you have to do here is to double the number of moles of iron(III) oxide and multiply the result by the molar mass of iron metal.

The reaction will produce

$12.52$ moles $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \frac{2 \text { moles } \mathrm{Fe}}{1 \text { mole-Fe } \mathrm{e}_{2} \mathrm{O}_{3}}=25.04$moles $\mathrm{Fe}$

This is equivalent to

$25.04 \text { moles } \mathrm{Fe} \cdot \frac{55.845 \mathrm{~g}}{1 \text { mole } \mathrm{Fe}}=\longdiv { 1 . 4 0 \cdot 1 0 ^ { 3 } \mathrm { g } }$

Final Answer:-

The answer must be rounded to three sig figs, the number of sig figs you have for the masses of the two reactants.

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