2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

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2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

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Answer 1 · 2015-01-31T09:26:13

Start by looking at the mole ratio between $C_{8} H_{18}$ $C_8H_18$ and $O_{2}$ $O_2$ ; notice that you need 25 moles of $O_{2}$ $"O_2$ for every 2 moles of $C_{8} H_{18}$ $C_8H_18$ . SInce you start with $4 moles$ $"4 moles"$ of $C_{8} H_{18}$ $C_8H_18$ , the number of $O_{2}$ $O_2$ moles you'll need is

${4.00 moles C}_{8} H_{18} \cdot \frac{{25 moles O}_{2}}{{2 moles C}_{8} H_{18}} = {50.0 moles O}_{2}$ $"4.00 moles C"_8"H"_18 * ("25 moles O"_2)/("2 moles C"_8"H"_18) = "50.0 moles O"_2$

Now all you need to do is use the ideal gas law equation, $PV = nRT$ $"PV" = "nRT"$ , to solve for the volume of $O_{2}$ $O_2$ needed (don't forget to transform degrees Celsius to Kelvin)

$P V = n R T \Rightarrow V = \frac{n R T}{P}$ $PV = nRT => V = (nRT)/P$

$V_{O_{2}} = \frac{50.0 moles \cdot 0.082 \frac{L \cdot atm}{mol \cdot K} \cdot (273.15 + 35) K}{0.953 atm}$ $V_(O_2) = ("50.0 moles" * "0.082" ("L" * "atm")/("mol" * "K") * ("273.15 + 35")"K")/("0.953 atm")$

$V_{O_{2}} = 1325.7 L$ $V_(O_2) = "1325.7 L"$ , or $V_{O_{2}} = 1330 L$ $V_(O_2) = "1330 L"$ - rounded to three sig figs.

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2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

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