If 8 g of hydrogen gas and 64 g of oxygen, gas or formed, then how many grams of water must have been used
To solve this problem, we need to determine the stoichiometric ratio between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O).
The balanced chemical equation for the reaction is:
2H2 + O2 -> 2H2O
The molar mass of hydrogen (H2) is 2 g/mol, and the molar mass of oxygen (O2) is 32 g/mol.
First, we convert the given masses of hydrogen and oxygen into moles:
Number of moles of H2 = 8 g / 2 g/mol = 4 mol
Number of moles of O2 = 64 g / 32 g/mol = 2 mol
From the balanced chemical equation, we see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. Therefore, the ratio is 2:1 for H2 to O2.
Since we have 4 moles of H2, we need half the number of moles of O2 (2 moles) to react fully with the hydrogen.
Finally, we convert the number of moles of O2 back into grams:
Mass of O2 = Number of moles of O2 * molar mass of O2
Mass of O2 = 2 mol * 32 g/mol = 64 g
Therefore, 64 grams of water must have been used.