Consider the following balanced equation for the combustion of methane, a compon

Question

Consider the following balanced equation for the combustion of methane, a component of natural gas. CH_4 (g) + 2 O_2(g) rightarrow CO_2 (g) + 2 H_2 O(g) Complete the following table, showing the appropriate masses of reactants and products. If the mass of a reactant is provided, fill in the mass of other reactants required to completely react with the given mass, as well as the mass of each product formed. If the mass of a product is lay provided, fill in the required masses of each reactant to make that amount of product, as well as the mass of the other product that is formed.

Explanation / Answer

CH4 + 2 O2 -----> CO2 + 2 H2O
number of moles of O2 = 2.37/32 = 0.0741
number of moles of CH4 = 0.0741/2 = 0.03705
mass of CH4 = 0.03705*16 = 0.5928 g
number of moles of CO2 = 0.0741/2 = 0.03705
mass of CO2 = 0.03705*44 = 1.6302 g
number of moles of H2O = 0.0741
mass of H2O = 0.0741*18 = 1.3338 g

number of moles of CH4 = 17.32/16 = 1.0825
number of moles of O2 = 2*1.0825 = 2.165
mass of O2 = 2.165*32 = 69.28 g
number of moles of CO2 = 1.0825
mass of CO2 = 1.0825*44 = 47.63 g
number of moles of H2O = 2*1.0825 = 2.165
mass of H2O = 2.165*18 = 38.97 g

number of moles of H2O = 15.32/18 = 0.8511
number of moles of CH4 = 0.8511/2 = 0.4255
mass of CH4 = 0.4255*16 = 6.808 g
number of moles of O2 = 0.8511
mass of O2 = 0.8511*32 = 27.235 g
number of moles of CO2 = 0.8511/2 = 0.4255
mass of CO2 = 0.4255*44 = 18.722 g

number of moles of CO2 = 2250/44 = 51.136 kmoles
number of moles of CH4 = 51.136 kmoles
mass of CH4 = 51.126 * 44 = 2249.54 kg
number of moles O2 = 2*51.136 = 102.27 kmoles
mass of O2 = 102.27 * 32 = 3272.64 kg
number of moles H2O = 102.27 k moles
mass of H2O = 102.27*18 = 1840.86 kg

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