Question

Methane and oxygen react to form carbon dioxide and water. what mass of water is formed if 3.2 g of methane reacts with 12.8 g of oxygen to produce 8.8 g of carbon dioxide?

Answer 1

According to the balanced equation, 1 mole of methane reacts with 2 moles of oxygen to produce 2 moles of water. We can calculate the mass of water formed by using the molar masses of the compounds involved.

Step-by-step explanation:

According to the balanced equation: CH4 + 2O2 → CO2 + 2H2O

We can see that 1 mole of methane reacts with 2 moles of oxygen to produce 2 moles of water. Using the molar masses of the compounds, we can calculate the mass of water formed.

The molar mass of water (H2O) is 18.016 g/mol. Therefore, the mass of 2 moles of water is (2 * 18.016) g/mol = 36.032 g/mol.

So, if 3.2 g of methane reacts with 12.8 g of oxygen to produce 8.8 g of carbon dioxide, the mass of water formed would be 36.032 g/mol - (8.8 g/mol + 44.01 g/mol) = 36.032 g/mol - 52.81 g/mol = -16.778 g/mol.

Answer 2

The chemical equation for the combustion reaction of methane and oxygen is,
CH₄ + 2O₂ --> CO₂ + 2H₂O

Note that the equation is already balanced which means that both sides will have the same number of atoms of each elements.

We use dimensional analysis and conversion to answer the question.

(1) 3.2 g methane
(3.2 g CH4)(1 mol CH4/16 g CH4)(2 mol H2O/1 mol CH4)(18 g H2O/1 mol H2O)
= 7.2 grams of H2O
(2) 12.8 g of oxygen
(12.8 g O2)(1 mol O2/16 g O2)(2 mol H2O/2 mol O2)(18 g H2O/1 mol H2O)
= 14.4 grams of H2O

(3) 8.8 g of carbon dioxide
(8.8 g CO2)(1 mol CO2/44 g CO2)(2 mol H2O/1 mol CO2)(18 g H2O/1 mol H2O)
= 7.2 grams of H2O

Since, the least number among the three answers is 7.2 grams. The answer is therefore, 7.2 grams of water.