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3 years ago

What mass of carbon dioxide gas occupies a volume of 81.3L at 204 kPa and a temperature of 95.0°C?

Chemistry

1 answer:

alexandr1967 [171]3 years ago

5 0

Answer:

236.9g

Explanation:

Given parameters:

Volume of gas = 81.3L

Pressure of gas = 204kPa

temperature of gas = 95°C

Unknown:

Mass of carbondioxide gas = ?

Solution:

To solve this problem, the ideal gas law will be well suited. The ideal gas law is a fusion of Boyle's law, Charles's law and Avogadro's law.

Mathematically, it is expressed as;

PV = nRT

the unknown here is n which is the number of moles;

P is the pressure, V is the volume, R is the gas constant and T is the temperature.

convert pressure into atm

101.325KPa = 1atm

204 kPa = $\frac{204}{101.325}$ = 2atm

Convert temperature to Kelvin; 95 + 273 = 368K

2 x 81.3 = n x 0.082 x 368

n = $\frac{2 x 81.3}{0.082 x 368}$ = 5.38moles

Since the unknown is mass;

Mass = number of moles x molar mass

Molar mass of carbon dioxide = 12 + 2(16) = 44g/mol

Mass = 5.38 x 44 = 236.9g

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Answer:

1. 15.71 g CO2

2. 38.19 % of efficiency

Explanation:

According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:

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CO = 28.01 g/mol

(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO

For CO2:

C = 12.01 g/mol

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We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:

(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2

We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:

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From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,

(6g / 15.71g) x 100 = 38.19 % of efficiency

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