how do you calculate the mass of water that will be produced when 5.0 grams of H2 is reacted with excess O2?
one O atom for every 2 H atoms
every H atom weighs 1 unit
every H2 molecule weighs 2 units
every O atom weighs 16 units
the sum of 2 units and 16 units is 18 units, the H2O mass
so
2/18 = Hydrogen mass / water mass = 1/9
so
(1/9) of our water mass is 5 grams
9*5 = 45 grams
To calculate the mass of water produced when 5.0 grams of H2 gas reacts with excess O2, you need to use the balanced chemical equation for the reaction and then perform a stoichiometric calculation.
The balanced chemical equation for the reaction between H2 and O2 to produce water (H2O) is:
2H2 + O2 -> 2H2O
From the equation, you can see that two moles of water are produced for every two moles of hydrogen gas (H2) reacted.
To find the number of moles of H2, you need to divide the given mass of H2 by its molar mass. The molar mass of hydrogen gas (H2) is 2.016 g/mol.
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 5.0 g / 2.016 g/mol = 2.48 mol
Since the stoichiometry of the reaction shows that 2 moles of H2 produce 2 moles of H2O, we can conclude that 2.48 moles of H2 will produce 2.48 moles of water.
Finally, to calculate the mass of water produced, you need to multiply the number of moles of water by its molar mass. The molar mass of water (H2O) is 18.015 g/mol.
Mass of water = Moles of water × Molar mass of water
Mass of water = 2.48 mol × 18.015 g/mol = 44.7 g
Therefore, when 5.0 grams of H2 reacts with excess O2, the mass of water produced is approximately 44.7 grams.