If 8 g of hydrogen gas and 64 g of oxygen gas performed how many grams of water must have been used
The given information suggests that a chemical reaction has occurred between hydrogen gas (H2) and oxygen gas (O2) to produce water (H2O).
To determine the amount of water produced, we need to first find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
We can follow these steps to find the limiting reactant and then calculate the amount of water produced:
1. Write the balanced chemical equation for the reaction:
2H2 + O2 -> 2H2O
2. Calculate the molar mass of each reactant:
Hydrogen gas (H2):
- Molar mass of H2 = 2 g/mol
Oxygen gas (O2):
- Molar mass of O2 = 32 g/mol
3. Convert the given masses of the reactants to moles:
Hydrogen gas (H2):
- Moles of H2 = mass / molar mass = 8 g / 2 g/mol = 4 mol
Oxygen gas (O2):
- Moles of O2 = mass / molar mass = 64 g / 32 g/mol = 2 mol
4. Determine the limiting reactant:
From the balanced chemical equation, we can see that the ratio of H2 to O2 is 2:1. This means that for every 2 moles of H2, we need 1 mole of O2. Since we have an equal number of moles of H2 (4 mol) and O2 (2 mol), the O2 is the limiting reactant in this case.
5. Calculate the moles of water formed using the stoichiometry from the balanced equation:
From the balanced chemical equation, we know that 2 moles of H2 react to form 2 moles of H2O. Therefore, for every 1 mole of O2 consumed, we will form 2 moles of H2O.
Moles of H2O = 2 * moles of O2 = 2 * 2 mol = 4 mol
6. Calculate the mass of water formed:
Molar mass of H2O = 18 g/mol
Mass of H2O = moles of H2O * molar mass of H2O = 4 mol * 18 g/mol = 72 g
Therefore, 72 grams of water must have been formed in this reaction.