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Nataly_w [17]
3 years ago
5

Calculate the volume of 4.50 mol of SO2 (g) measured at STP. At STP: P = 101.3 kPa, T = 273.15 K A. 87.3 L B. 101 L C. 127 L D.

232 L
Chemistry
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

B

101L

Explanation:

We use the ideal gas relation

PV = nRT

P = pressure = 101.3KPa

V = volume = ?

n = number of moles = 4.5moles

T = Temperature = 273.15K

R = molar gas constant = 8.314J/mol.k

Rearranging the equation to make V the subject of the formula yields :

V = nRT/P

= ( 4.5 × 8.314 × 273.15) ÷ 101.3

= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L

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<h3>Answer:</h3>

Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

<h3>Explanation:</h3>

From the questions we are given;

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(s)

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We are supposed to determine the limiting and excess reactants;

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Working with the amount of reactants given;

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Conclusion;

We can conclude that Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

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