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Misha Larkins [42]
3 years ago
11

using the equation, c5h12 8o2 arrow 5co2 6h2o, if 108 g of water are produced, how many grams of oxygen were consumed?

Chemistry
1 answer:
vfiekz [6]3 years ago
8 0
Molar mass:

H₂O = 18.0 g/mol

O₂ = 32.0 g/mol

C₅H₁₂ + 8 O₂ -> 5 CO₂ + 6 H₂<span>O
</span>
8 x (32 g )<span> ------------ 6 x (18 g )</span>
mass O₂ ------------ 108 g H₂O

mass O₂ = 108 x 8 x 32 / 6 x 18

mass O₂ = 27648 / 108

mass O₂ =<span> 256 g</span>

<span>hope this helps!</span>
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<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

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W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

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