Question

How many grams of oxygen react with 72.0 grams of C5H12 by using this equation-C5H12+8O2 and 5CO2+6H2O?

Answer 1

Answer on Question #65317 - Chemistry – Other

Task:

How many grams of oxygen react with 72.0 grams of $C_5H_{12}$ by using this equation:

$C_5H_{12} + 8O_2 = 5CO_2 + 6H_2O$?

Solution:

Reaction equation:

$$C_5H_{12} + 8O_2 = 5CO_2 + 6H_2O$$

We find the amount of pentane ($C_5H_{12}$):

$$M(C_5H_{12}) = 5 \times Ar(C) + 12 \times Ar(H) = 5 \times 12 + 12 \times 1 = 60 + 12 = 72 \text{ (g/mol)};$$ $$n(C_5H_{12}) = \frac{m(C_5H_{12})}{M(C_5H_{12})} = \frac{72.0 \text{ g}}{72 \text{ g/mol}} = 1 \text{ mol of } C_5H_{12}.$$

72 grams of pentane = 1 mol

By reaction equation:

$$n(C_5H_{12}) = \frac{n(O_2)}{8}; \Rightarrow n(O_2) = 8 \times n(C_5H_{12});$$

We find the mass of oxygen ($O_2$):

$$M(O_2) = 2 \times Ar(O) = 2 \times 16 = 32 \text{ (g/mol)};$$ $$m(O_2) = n(O_2) \times M(O_2) = 8 \times n(C_5H_{12}) \times M(O_2) = 8 \times 1 \times 32 = 256 \text{ grams of } O_2$$

Answer: 256 grams of oxygen react with 72.0 grams of $C_5H_{12}$.

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