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3 years ago

What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3

2 answers:

6 0

First we will calculate the number of moles of Iron:
$n = \frac{m}{M}
$ , where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
$n= \frac{84.9}{55.845} = 1,52$ moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So $1,52= \frac{mOxygen}{32}$ where 32 is the Atomical Weight of Oxygen (16 x 2).
=> $mOxygen=32*1,52=48,64$ g

saul85 [17]3 years ago

3 0

Answer:

36.385 grams of oxygen reacts when 84.9 grams of iron.

Explanation:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Moles of iron = $\frac{84.9 g}{56 g/mol}=1.5160 mol$

According to reaction, 4 moles of iron reacts with 3 moles of oxygen gas.

Then 1.5160 moles of iron will react with:

$\frac{3}{4}\times 1.5160 mol=1.1370 mol$ of oxygen gas

Mass of 1.1370 moles of oxygen gas:

$1.1370 mol\times 32 g/mol=36.385 g$

36.385 grams of oxygen reacts when 84.9 grams of iron.

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

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$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

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Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

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$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

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