Question

Methane, $ C{H_4} $ , the major component of natural gas, burns in air to form $ C{O_2} $ and $ {H_2}O $ . What mass of water is formed in the complete combustion of $ 5.00 \times {10^3}g $ of $ C{H_4} $ ?

Answer 1

Hint: The moles of methane can be determined from the molar mass and given mass of methane. From the balanced chemical equation of combustion of methane, the moles of water can be determined as one mole of methane gives two moles of water. The moles and molar mass of water gives mass of water.

Complete Step By Step Answer:
Combustion reaction is the reaction between alkanes when treated with oxygen gas leads to the formation of carbon dioxide and water.
Given that methane is a major component of combustion.
The mass of methane is given as $ 5.00 \times {10^3}g $ and the molar mass of methane is $ 16gmo{l^{ - 1}} $
Thus, moles of methane is $ {n_{MeOH}} = \dfrac{{5000}}{{16}} = 312.5moles $
The combustion reaction of methane is
 $ C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O $
From the balanced combustion reaction of methane, one mole of methane produces two moles of water.
Moles of water is $ {n_{{H_2}O}} = 312.5 \times 2 = 625moles $
The molar mass of water is $ 18gmo{l^{ - 1}} $
Thus, mass of water is the product of molar mass and moles
 $ {m_{{H_2}O}} = 625 \times 18 = 11250g $
By converting the obtained mass of water into kilograms, $ 11250g = 11.2kg $
Thus, the mass of water is $ 11.2kg $ .

Note:
The number of moles of a compound is the ratio of the mass to molar mass which has the units of mole. The mass is given in grams, and the molar mass is in grams per mole. But the final answer is in kilograms but not in grams. Thus, conversion must be done from grams to kilograms.