Question

In the reaction c4h10 + o2= co2 + h2o 172.5 g of c4h10 reacts with excess o2. How many grams of co2 will be produced?

Answer 1

Answer on Question #83621 – Chemistry – Organic Chemistry

Task:

In the reaction $C_4H_{10} + O_2 = CO_2 + H_2O$, 172.5 g of $C_4H_{10}$ reacts with excess $O_2$. How many grams of $CO_2$ will be produced?

Solution:

Chemical reaction equation:

$$2 C _ {4} H _ {1 0} + 1 3 O _ {2} = 8 C O _ {2} + 1 0 H _ {2} O$$

According to the chemical reaction equation:

$$\frac {n (C _ {4} H _ {1 0})}{2} = \frac {n (C O _ {2})}{8}; \quad \Rightarrow \quad n (C _ {4} H _ {1 0}) = \frac {n (C O _ {2})}{4}$$

Then,

$$M \left(C _ {4} H _ {1 0}\right) = 4 * A r (C) + 1 0 * A r (H) = 4 * 1 2 + 1 0 * 1 = 5 8 \frac {g}{m o l};$$ $$M \left(C O _ {2}\right) = A r (C) + 2 * A r (O) = 1 2 + 2 * 1 6 = 4 4 \frac {g}{m o l};$$ $$n (X) = \frac {m (X)}{M (X)};$$ $$\frac {m (C _ {4} H _ {1 0})}{M (C _ {4} H _ {1 0})} = \frac {m (C O _ {2})}{4 * M (C O _ {2})};$$ $$m (C O _ {2}) = \frac {4 * M (C O _ {2}) * m (C _ {4} H _ {1 0})}{M (C _ {4} H _ {1 0})};$$ $$m (C O _ {2}) = \frac {4 * 4 4 \frac {g}{m o l} * 1 7 2 . 5 g}{5 8 \frac {g}{m o l}} = 5 2 3. 4 4 8 g;$$ $$m \left(C O _ {2}\right) = 5 2 3. 4 4 8 g$$

Answer: 523.448 grams of $CO_2$ will be produced.

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