Question

Determina a quantidade da agua produzida quando 20 gramas de H2 reage com 8 gramas de O2? a) Indique o reagente limitante ? b) Determinar a quantidade em excesso

Answer 1

Answer:

1. Produced mass of water: $m_{H_2O}=9gH_2O$

2. Limiting reactant: oxygen.

3. Excess hydrogen: $m_{H_2}^{excess}=19gH_2$

Explanation:

Hello,

In this case, given the reaction:

$2H_2+O_2\rightarrow 2H_2O$

a) We identify the limiting reactant by firstly computing the available moles of hydrogen and the moles of hydrogen actually consumed by the given 8 g of oxygen a shown below:

$n_{H_2}^{available}=20gH_2\times\frac{1molH_2}{2gH_2}=10molH_2\\\\n_{H_2}^{\consumed\ by\ O_2}=8gO_2*\frac{1molO_2}{32gO_2}*\frac{2molH_2}{1molO_2} =0.5molH_2$

Thus, since there are more available hydrogen than it that is reacting, we conclude it is in excess, consequently, oxygen is the limiting reactant. Moreover, we compute the produced grams of water for the consumed 0.5 mol of hydrogen by oxygen as shown below:

$m_{H_2O}=0.5molH_2*\frac{2molH_2O}{2molH_2} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=9gH_2O$

(b) Finally, the excess hydrogen is:

$m_{H_2}^{excess}=(10molH_2-0.5molH_2)\times \frac{2gH_2}{1molH_2} \\\\m_{H_2}^{excess}=19gH_2$

Best regards.