Fe2O3+2Al=Al2O3+2Fe A welder has 1.873 × 10^2 g Fe2O3 and 94.51 g Al in his welding kit. Which reactant will he run out of first? (options: Fe2O3 or Al.......it's Fe2O3). How much of this reactant should he order to make sure he runs out of both reactants at the same time? He should order ___g of the limiting reactant.

Answers

Answer 1
Answer:

Answer: a) Fe_2O_3

b) 92.1 g

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of Fe_2O_3

\text{Number of moles}=(1.873* 10^2 g)/(159.69g/mol)=1.173moles

b) moles of Al

\text{Number of moles}=(94.51g)/(27g/mol)=3.500moles

Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe

According to stoichiometry :

1 mole of Fe_2O_3 require 2 moles of Al

Thus 1.173 Cl_2 require=(2)/(1)* 1.173=2.346moles  of Al

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product. Thus Fe_2O_3 will run out first.

Al is the excess reagent as (3.500-2.346)= 1.154 moles are left unreacted.

2 moles of Al require 1 mole of Fe_2O_3

3.500 moles of Al require= (1)/(2)* 3.500=1.750moles of Fe_2O_3

Moles of Fe_2O_3  required = (1.750-1.173) = 0.577

Mass of Fe_2O_3=moles* {\text {Molar mass}}=0.577moles* 159.69g/mol=92.1g

Thus he should order 92.1 g of the limiting reactant.

Answer 2
Answer:

Answer:

92.41g

Explanation:


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Answers

Answer:

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Final answer:

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Explanation:

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Answers

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Answers

Answer:

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Explanation:

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Answers

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