When methane (ch4) is burned with oxygen, the products are carbon dioxide and water. if you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?

When methane (CH4) is burned with oxygen, 4 g of methane were needed for the reaction.

The equation of the reaction is;

CH4(g) + 2O2(g) > CO2(g) + 2H2O(g)

Number of moles in 11g of CO2 = 11g/44g/mol = 0.25 moles of CO2

Number of moles in 16 g of O2 = 16 g/32g/mol = 0.5 moles

Since the reaction is 2:1

2 mole of O2 yields 1 mole of CO2

x moles of O2 yields 0.25 moles of CO2

x = 2  mole *  0.25 moles/1 mole

= 0.5 moles of O2

Also, the reaction between methane and oxygen is 1:2

1 mole of CH4 reacts with 2 moles of oxygen

x moles of methane reacts with 0.5 moles of O2

x = 1 mole *  0.5 moles/ 2 moles

x =  0.25 moles of CH4

Mass of 0.25 moles of CH4 = 0.25 moles *  16 g/mol = 4 g of CH4

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the answer is 0.42 is that is an option. this question provides no info