Calculate the mass of CO needed to react completely with 50.Og of Fe2 03?

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Calculate the mass of CO needed to react completely with 50.Og of Fe2 03?

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Answer 1 · 2018-04-06T12:42:21

$\approx 26.31$ $approx 26.31$ grams

Explanation:

We must first create a balanced equation. Carbon monoxide is commonly used in industry to liberate iron from its oxides (which are often found in rock). This produced Carbon dioxide and solid iron.

$F e_{2} O_{3} (s) + 3 C O (g) \to 2 F e (s) + 3 C O_{2} (g)$ $Fe_2O_3(s)+3CO(g) ->2Fe(s) +3CO_2(g)$

As $C O$ $CO$ is used to liberate iron from Iron (III) Oxide.

Now we must find how many moles 50.0 g of Iron (III) oxide corresponds to:

$M$ $M$ (iron (III) oxide)= $159.69 g m o l^{- 1}$ $159.69 gmol^-1$
$m = 50.0$ $m=50.0$
$n$ $n$ =?

Using the equation:

$n = \frac{m}{M}$ $n=(m)/(M)$

$n = \frac{50.0}{159.69}$ $n=(50.0)/(159.69)$

$n \approx 0.3131$ $n approx 0.3131$

Now from the molar ratios we can see that we require 3 moles of $C O$ $CO$ for every 1 mole of Iron (III) oxide.

$n (C O) = 3 \times 0.3131 = 0.9393$ $n(CO)=3 times 0.3131=0.9393$

Then using the same equation we can find the mass of $C O$ $CO$ required:

$n = \frac{m}{M}$ $n=(m)/(M)$
$M (C O) = 28.01 g m o l^{- 1}$ $M(CO)=28.01 gmol^-1$

$0.9393 = \frac{m}{28.01}$ $0.9393=(m)/(28.01)$

$m \approx 26.31 g$ $m approx 26.31 g$ of $C O$ $CO$ is needed.

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Calculate the mass of CO needed to react completely with 50.Og of Fe2 03?

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