Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.481 g of methane is mixed with 0.54 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

From the reaction equation, we can see that one mole of methane gave 2 moles of water. This is the theoretical yield. We need to note the actual yield.

To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.

Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g

Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875

Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.

The mass is thus, 0.016875 * 18 = 0.30375g

1.08g is higher and thus is the maximum mass