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.

Question

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.

jhannybean · Answer

irst convert the number of grams of each to moles. $$\sf 1.873 ~x~ 10^2 \text{g}~ Fe_2O_3\cdot \frac{1~\text{mol}~Fe_2O_3}{159.7~\text{g}~Fe_2O_3} =~?~ mol~Fe_2O_3 ~(have) $$$$\sf 94.51~g~Al \cdot \frac{1 ~mol ~Al}{26.98~g ~Al}=~? ~mol~Al~(have)$$Then pick one of the two to do a mole to mole ratio. $$\sf ~?~mol~Al \cdot \frac{1~mol~Fe_2O_3}{2~mol~Al}=~?~\text{mol of}~~Fe_2O_2 ~is~needed~$$If the amount needed is bigger than the amount you have, then the element or compound you used to find the mole ratio is the LR. If the moles of $\sf Fe_2O_3$ needed is less than what you have, then that element or compound is the excess.

anonymous · Answer

Thank you soooo much!!!!