Question #4187f

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Question #4187f

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Answer 1 · 2017-05-03T08:17:05

120 mL

Explanation:

$n m o l C_{6} H_{6} = \frac{39 g}{78 \frac{g}{m o l}}$ $n mol C_6H_6 = (39g)/(78g/(mol)$ = $0.4997 m o l$ $0.4997mol$

$\frac{15 m o l O_{2}}{2 m o l C_{6} H_{6}} \times 0.4997 m o l C_{6} H_{6}$ $(15mol O_2)/(2 mol C_6H_6) times 0.4997 mol C_6H_6$ = $3.784 m o l O_{2}$ $3.784mol O_2$

$G r a m s O_{2} = m a s s O_{2} = n \times M W O_{2} = 3.784 \times 32 = 119.9 g$ $Grams O_2 = mass O_2 = n times MW O_2 = 3.784 times 32 = 119.9g$
Density liquid $O_2 = 1.1 g/(mL), therefore,$ amount in grams = 119.9

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Question #4187f

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