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3 years ago

What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O

Chemistry

1 answer:

Ipatiy [6.2K]3 years ago

6 0

Answer:

Mass = 13.23 g

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

O₂ : C₃H₈

5 : 1

1.5 : 1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g

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To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + 2 NH₄SCN(s) → Ba(SCN)₂(s) +10 H₂O(l) + 2 NH₃(g)

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It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

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m = molality of solute (m or $\frac{mol}{Kg}$ )

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Plugging value in expression :

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Hence change in freezing point = 1.69 °C

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