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Example 13:0. 29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide at S. T. P. From the given information, calculate the (i) mass of carbon dioxide formed. ​

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Answer:

0.779

Step-by-step explanation:

Determine the molecular weight of the hydrocarbon. We know that its vapor density is 29, which means that one mole of the hydrocarbon has a mass of 29 grams. Therefore, the molecular weight of the hydrocarbon is 29 g/mol.

Calculate the number of moles of the hydrocarbon. We can use the formula:

moles = mass / molecular weight

Substituting the values, we get:

moles = 29 g / 29 g/mol = 1 mol

Therefore, we have one mole of the hydrocarbon.

Write the balanced chemical equation for the combustion of the hydrocarbon in oxygen. The general equation is:

hydrocarbon + oxygen → carbon dioxide + water

For one mole of the hydrocarbon, we need one mole of oxygen to completely burn it. The balanced equation is:

CnHm + (n+m/4) O2 → n CO2 + m/2 H2O

Calculate the volume of carbon dioxide produced. We know that 1 mole of any gas at STP occupies 22.4 L. Therefore, one mole of carbon dioxide occupies 22.4 L. The volume of 448 ml of carbon dioxide at STP can be converted to liters:

448 ml = 0.448 L

The number of moles of carbon dioxide produced can be calculated using the ideal gas law:

PV = nRT

where P is the pressure (1 atm), V is the volume (0.448 L), n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature (273 K). Substituting the values, we get:

n = PV/RT = (1 atm x 0.448 L) / (0.0821 L atm/mol K x 273 K) = 0.0177 mol

Therefore, 0.0177 moles of carbon dioxide are produced.

Calculate the mass of carbon dioxide produced. We can use the formula:

mass = moles x molecular weight

The molecular weight of carbon dioxide is 44 g/mol. Substituting the values, we get:

mass = 0.0177 mol x 44 g/mol = 0.779 g

Therefore, the mass of carbon dioxide produced is 0.779 grams.

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User Rohan Sharma
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