Answer:
[tex]\boxed{\text{22.88 g}}[/tex]
Explanation:
This looks like a limiting reactant problem.
We are given the masses of two reactants and asked to determine if we have enough to make a given amount of product.
1. Assemble the information
We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 2.016 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
Mass/g: 2.56 20.32
2. Calculate the moles of each reactant
[tex]\text{Moles of H$_{2}$} = \text{2.56 g H}_{2} \times \dfrac{\text{1 mol H$_{2}$}}{\text{2.016 g H$_{2}$}} = \text{1.270 mol H$_{2}$}\\\\\text{Moles of O$_{2}$} = \text{20.32 g O}_{2} \times \dfrac{\text{1 mol O$_{2}$}}{\text{32.00 g O$_{2}$}} = \text{0.6350 mol O$_{2}$}[/tex]
3. Calculate the moles of H₂O from each reactant
[tex]\textbf{From H$_{2}$:}\\\text{Moles of H$_{2}$O} = \text{1.270 mol H$_{2}$} \times \dfrac{\text{2 mol H}_{2}\text{O}}{\text{2 mol H$_{2}$}} = \text{1.270 mol H$_{2}$O}\\\textbf{From O$_{2}$:}\\\text{Moles of H$_{2}$O} =\text{0.6350 mol O$_{2}$} \times \dfrac{\text{2 mol H}_{2}\text{O}}{\text{1 mol O$_{2}$}} = \text{1.270 mol H$_{2}$}\text{O}\\\text{You have just enough hydrogen and oxygen to react with each other.}\\\text{There is no limiting reactant.}[/tex]
4. Calculate the mass of H₂O
[tex]\text{Mass of H$_{2}$O} = \text{Mass of reactants} =\text{2.56 g + 20.32 g } = \textbf{22.88 g}\\\\\text{The mass of water formed is $\boxed{\textbf{22.88 g}}$}[/tex]
