Question

Сколько литров воздуха потребуется для сжигания 32 грамм метана??

Answer 1

Дано:
m(CH4) = 32 г

Найти: 
V(возд.) - ? (л)

Решение:
СH4 + 2O2 ---> CO2 + 2H2O 
n(CH4) = 32/16 = 2 моль
n(O2) = 2n(CH4) = 4 моль 
V(O2) = 22,4 x 4 = 89,6 л

Ответ: V(O2) = 89,6 л.

Answer 2

Calculating Air Needed to Burn Methane

To calculate the amount of air needed to burn 32 grams of methane, we can use the stoichiometric ratio of methane to oxygen in the combustion reaction.

The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

This equation shows that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

Calculation

1. Convert grams of methane to moles: - The molar mass of methane (CH4) is approximately 16 grams/mol. - Therefore, 32 grams of methane is equal to 32/16 = 2 moles of methane.

2. Calculate the moles of oxygen needed: - According to the balanced chemical equation, 1 mole of methane requires 2 moles of oxygen. - So, 2 moles of methane will require 2 * 2 = 4 moles of oxygen.

3. Convert moles of oxygen to liters of air at STP (Standard Temperature and Pressure): - At STP, 1 mole of any gas occupies 22.4 liters. - Therefore, 4 moles of oxygen will occupy 4 * 22.4 = 89.6 liters of air.

Answer

To burn 32 grams of methane, approximately 89.6 liters of air will be needed.