I apologize, there was an error in my previous response. Let's try solving the problem again:
We know that 8 grams of hydrogen gas (H2) and 64 grams of oxygen gas (O2) are formed.
To calculate how many grams of water (H2O) must have been used, we need to recognize that the mass of the reactants must equal the mass of the products in a chemical reaction.
The equation for the formation of water is:
2H2 + O2 -> 2H2O
To find the mass of water used, we can subtract the mass of hydrogen and oxygen gas formed from the total mass of water formed:
Mass of water used = Mass of water formed - (Mass of hydrogen gas formed + Mass of oxygen gas formed)
Since the molar mass of hydrogen is 1 g/mol and the molar mass of oxygen is 16 g/mol, we can calculate the molar mass of water (H2O):
Molar mass of water (H2O) = (2 * molar mass of hydrogen) + molar mass of oxygen
= (2 * 1 g/mol) + 16 g/mol
= 2 g/mol + 16 g/mol
= 18 g/mol
Now, let's calculate the number of moles of hydrogen gas formed:
Moles of hydrogen gas formed = Mass of hydrogen gas formed / molar mass of hydrogen
= 8 g / 2 g/mol
= 4 mol
Similarly, let's calculate the number of moles of oxygen gas formed:
Moles of oxygen gas formed = Mass of oxygen gas formed / molar mass of oxygen
= 64 g / 32 g/mol
= 2 mol
Finally, we can calculate the mass of water used:
Mass of water used = (2 * moles of water formed) - (moles of hydrogen gas formed * molar mass of water) - (moles of oxygen gas formed * molar mass of water)
= (2 * 4 mol) - (4 mol * 18 g/mol) - (2 mol * 18 g/mol)
= 8 mol - 72 g - 36 g
= 8 mol - 108 g
= -100 g
Based on this calculation, it appears that there is a discrepancy. The mass of water used is -100 grams, which does not make sense in this context. It seems there may be an error in the given information or calculation.