Question

Calculate the volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

Answer 1

535.8 L volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given:

$C(s) + O_2(g)$ ⇒$CO_2(g)$

1 mole of $CO_2(g)$ is formed from 1 mole of charcoal, that means 1 mole of $CO_2(g)$ is formed from 12 g of charcoal.

1 lbs = 453.592 grams

6.75 lbs = 6.75 x 453.592 =3061.746g

So, $CO_2(g)$ formed from 3061.746g of charcoal is:

=$\frac{1}{12}$ x 3061.746g

=255.14 mol.

Now, at STP, the volume of 1 mol of gas =22.4 L

Volume of 255.14 mol of $CO_2(g)$ = 255.14 x 22.4 L

Volume of 255.14 mol of $CO_2(g)$ = 535.8 L

Hence, 535.8 L volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

Learn more about the ideal gas here:

brainly.com/question/27691721

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