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Amanda [17]
3 years ago
12

C3H8+5O2 →3CO2+4H20,

Chemistry
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

6.63 grams of CO₂ can be produced.

Explanation:

1 mol of C₃H₈ reacts to 5 moles of oxygen, in a combustion reaction in order to produce 3 moles of carbon dioxide and 4 moles of water.

First of all, we need to know, the limiting reactant:

5.52 g . 1mol / 44g = 0.125 moles of C₃H₈

8.03 g . 1mol/ 32g = 0.251 moles of oxygen.

1 mol of C₃H₈ can react to 5 moles of O₂

Then, 0.125 moles, must react to (0.125 . 5) /1 = 0.625 moles.

We only have 0.251 moles of O₂ and we need 0.625 moles. Cause we do not have enough oxygen, that's the limiting reagent.

5 moles of oxygen can produce 3 moles of CO₂

0.251 moles may produce (0.251 . 3) /5 = 0.1506 moles.

We convert the moles to mass : 0.1506 mol . 44g /mol = 6.63 g

Reaction is: C₃H₈  +  5O₂ →  3CO₂  + 4H₂O

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A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
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mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

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3 years ago
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