2 Answers

Bjarni Ragnarsson · Answer 1 · 2022-08-02T16:48:00+0000

83.3% yield

Step-by-step explanation:

First, we need to convert 240 g of
Fe_(2)O_(3) into moles:

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

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

Next, find the theoretical Fe yield using molar ratios.

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

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

Then convert this back into grams:

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

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

$\frac{140\:\text{g}\:Fe}{168\:\text{g}\:Fe} * 100$ %= 83.3%

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