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Bumek [7]
3 years ago
9

What mass of CaCO3 is required to react completely with 0.56 L of HCl?

Chemistry
1 answer:
Lina20 [59]3 years ago
8 0
The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol 
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g 
1.25 g of CaCO₃ is required 
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If I need 2.2 moles of CO2 , and I have excess Fe2O3 , how many moles of C do I need?
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Answer:

0.733 mol.

Explanation:

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<em>2Fe₂O₃ + C → Fe + 3CO₂,</em>

It is clear that 1.0 moles of Fe₂O₃  react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.

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<u><em>Using cross multiplication:</em></u>

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??? mole of C produces → 2.2 moles of CO₂.

∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.

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