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nadezda [96]
3 years ago
10

How many grams of gas must be released from a 32.0 L sample of CO2(g) at STP to reduce the volume to 16.6 L at STP?

Chemistry
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

30.3 g

Explanation:

At STP, 1 mol of any gas will occupy 22.4 L.

With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:

  • 32.0 L ÷ 22.4 L/mol = 1.43 mol

Then we <u>calculate how many moles would there be in 16.6 L</u>:

  • 16.6 L ÷ 22.4 L/mol = 0.741 mol

The <u>difference in moles is</u>:

  • 1.43 mol - 0.741 mol = 0.689 mol

Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 0.689 mol * 44 g/mol = 30.3 g
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Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one
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The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

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<em />

The mass of the octane burnt is calculated as follows;

m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g

The combustion reaction of octane is given as;

2C_8H_{18} +  \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O

From the reaction above:

228.46 g of octane -------------------> 704 g of  CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

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