Question

the process of producing gasoline from coal includes a gasification step to make hydrogen or synthesis gas. if 50kg test run of gas gives the average of 10% 40%CH4, 30% CO, and 20% CO2. What is the average molecular weight of the gas?

Answer 1

Answer on Question#65104 – Chemistry – Organic chemistry

Question: the process of producing gasoline from coal includes a gasification step to make hydrogen or synthesis gas. if 50kg test run of gas gives the average of 10% 40%CH4, 30% CO, and 20% CO2. What is the average molecular weight of the gas?

Solution:

1) $m_{i} = w_{i} \times m_{gas}$

A hydrogen is a missed component.

$$m_{H_2} = 0.1 \times 50kg = 5kg = 5 \times 10^3g$$ $$m_{CH_4} = 0.4 \times 50kg = 20kg = 2 \times 10^4g$$ $$m_{CO} = 0.3 \times 50kg = 15kg = 1.5 \times 10^4g$$ $$m_{CO_2} = 0.2 \times 50kg = 10kg = 1 \times 10^4g$$

2) $n_i = \frac{m_i}{M_i}$

$$n_{H_2} = \frac{5 \times 10^3g}{2 \frac{g}{mol}} = 2.5 \times 10^3mol = 2500\ mol$$ $$n_{CH_4} = \frac{2 \times 10^4g}{16 \frac{g}{mol}} = 1.25 \times 10^3mol = 1250\ mol$$ $$n_{CO} = \frac{1.5 \times 10^4g}{28 \frac{g}{mol}} = 536\ mol$$ $$n_{CO_2} = \frac{1 \times 10^4g}{44 \frac{g}{mol}} = 227\ mol$$ $$\sum n = 2500\ mol + 1250\ mol + 536\ mol + 227\ mol = 4513\ mol$$ $$\bar{M} = \frac{m_{gas}}{\sum n} = \frac{5000g}{4513\ mol} = 11 \frac{g}{mol}$$

Answer:

$$11 \frac{g}{mol}.$$

Answer provided by AssignmentExpert.com