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3 years ago

Mohamedaidh00 r u there pls reply

Chemistry

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professor190 [17]3 years ago

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Answer:

yess I'm here ............

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How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.

Fed [463]

Explanation:

$3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)$

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of $Fe$ in 210.3 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{210.3}{55.84 g/mol}=3.76 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : $\frac{3}{2}\times 3.76 moles$ of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g$

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of $Fe$ in 209.7 g=

$\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}$

$=\frac{209.7}{55.84 g/mol}=3.75 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : $\frac{3}{2}\times 3.75 moles$ of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

$\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g$

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Natasha2012 [34]

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Mohamedaidh00 r u there pls reply​

Mohamedaidh00 r u there pls reply