Question

If you need 66 grams of carbon dioxide. How many liters of propane do you need?

Answer 1

The question requires us to calculate the amount of propane, in liters, required to obtain 66g of carbon dioxide.

Considering the combustion of propane (C3H8) as the reaction to obtain carbon dioxide (CO2), we have the chemical reaction as follows:

`2C_3H_(8(l))+9O_(2(g))\rightarrow4CO_(2(g))+2CO_((g))+8H_2O_((v))`

From the reaction we can estabilish the following stoichiometric relationship between propane and carbon dioxide:

2 mol C3H8 ---------- 4 mol CO2

Now, we need to calculate the amount of C3H8, in moles, necessary to obtain 66g of CO2, knowing that the molar mass of CO2 is 44g/mol:

2 mol C3H8 ---------- 4 mol CO2

x ----------- (66g CO2 / 44g/mol CO2) = 1.5 mol CO2

Solving for x, we have that 0.75 moles of C3H8 are required to produce 66g of CO2 (note that we had to convert the mass of CO2, 66g, into its number of moles, 1.5 mol to make this calculations).

Next, considering the Standard Temperature and Pressure conditions (STP), where 1 mol of a gas corresponds to 22.4 L of this gas, we can calculate the volume of C3H8 necessary to produce 66g of CO2:

1 mol C3H8 ---------- 22.4 L of C3H8

0.75 mol C3H8 ---- y

Solving for y, we have that 16.8 L of propane are required to obtain 66g of CO2.