Question

Can someone please solve this for me and explain it

Answer 1

83.3% yield

Explanation:

First, we need to convert 240 g of $Fe_{2}O_{3}$ into moles:

$240 \:g \:Fe_{2}O_{3} \:\times(\frac{1\:\text{mol}\:Fe_{2}O_{3}}{159.69\: \text {g}\:Fe_{2}O_{3}})$

$=1.50 \:\text{mol}\:Fe_{2}O_{3}$

Next, find the theoretical Fe yield using molar ratios.

$1.50 \: \text {mol} \: Fe_{2}O_{3}\: \times (\frac{2\: \text{mol} \: Fe}{1 \:\text{mol} \: Fe_{2}O_{3}})$

$= 3.00 \: \text{mol} \: Fe$

Then convert this back into grams:

$3.00 \: \text{mol} \:Fe \times (\frac{55.845 \: \text{g} \: Fe}{1 \: \text{mol} \: Fe}) = 168 \: \text{g} \: Fe$

Note that actual yield is only 140 g Fe so percentage yield is

$\dfrac{140\:\text{g}\:Fe}{168\:\text{g}\:Fe} \times 100$%= 83.3%

Answer 2

83.3%

The answer is 83.3%

BRAINILIEST PLEASE