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dedylja [7]
2 years ago
11

Iron (III) oxide reacts with carbon monoxide producing iron metal and carbon dioxide. How much grams of iron is produced from 74

9 g of iron (III) oxide
Chemistry
1 answer:
Katarina [22]2 years ago
3 0

Answer:

389.48g of Fe

Explanation:

This is a stoichiometric reaction. First write a balanced equation of reaction: Fe2O3+3CO==>2Fe+3CO2

Since Fe2O3 and Fe are concerned with, calculate there masses in the balanced equation

Taking Molar mass of Fe and O to be 26 and 16 respectively.

Fe2O3 = (2×26) + (3×16)

=100g/mol

2Fe = 2×26

= 52

Therefore 100g of Fe2O3 gave 52g

749g of Fe2O3 will give

749×52/100

= 389.48g of Fe

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What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are
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54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

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Mole of CaCO3 =?

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Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

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This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

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