Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water . Suppose 2.73 g of methane is mixed with 6.7 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

3.8g of H2O are produced

Explanation:

The balanced reaction of the problem is:

CH4 + 2O2 → CO2 + 2H2O

Where 1 mole of CH4 reacts with 2 moles of O2

To solve this question we need to find, as first, the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles of H2O produced and its mass as follows:

Moles CH4 - 16.04g/mol-

2.73g * (1mol/16.04g) = 0.170 moles CH4

Moles O2 -32g/mol-

6.7g (1mol/32g) = 0.209 moles O2

For a complete reaction of 0.170 moles of CH4 are needed:

0.170 moles CH4 * (2 mol O2 / 1mol CH4) = 0.340 moles O2

As there are just 0.209 moles of O2, oxygen is limiting reactant

The moles of water produced are:

0.209 moles O2 * (2mol H2O / 2mol O2) = 0.209 moles H2O

Mass water -Molar mass: 18.01g/mol-

0.209 moles H2O * (18.01g/mol) = 3.8g of H2O are produced

pie chart because it is in percents ( % )