Question #12a91

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Question #12a91

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Answer 1 · 2017-05-16T01:49:52

$2.07 g C_{3} H_{8}$ $2.07gC_3H_8$

Explanation:

First, let's find the moles of $O_{2}$ $O_2$ using the ideal-gas equation:

$m o l O_{2} = \frac{(1.04 a t m) (5.53 L)}{(0.08206 \frac{L - a t m}{m o l - K}) (298 K)} = 0.235 m o l O_{2}$ $molO_2 = ((1.04atm)(5.53L))/((0.08206 (L-atm)/(mol-K))(298K)) = 0.235 molO_2$

Now, we'll use the stoichiometrically equivalent values from the equation to find the relative number of moles of $C_{3} H_{8}$ $C_3H_8$ :

$0.235 m o l O_{2} (\frac{1 m o l C_{3} H_{8}}{5 m o l O_{2}}) = 0.0470 m o l C_{3} H_{8}$ $0.235molO_2((1 molC_3H_8)/(5 mol O_2)) = 0.0470molC_3H_8$

Finally, we'll use the molar mass of propane to find the mass in grams of propane needed:

$0.0470 m o l C_{3} H_{8} (\frac{44.11 g C_{3} H_{8}}{1 m o l C_{3} H_{8}}) = 2.07 g C_{3} H_{8}$ $0.0470molC_3H_8((44.11gC_3H_8)/(1 mol C_3H_8)) = 2.07gC_3H_8$

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Question #12a91

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