Respuesta :
ANSWER
The mass of H2O that reacted is 65.88 grams
EXPLANATION
Given that;
The volume of oxygen atom at STP is 40.9L
Follow the steps below to find the mass of water reacted
Step 1; Write the balanced equation of the reaction
[tex]\text{ 2H}_2O_{(g)}\text{ }\rightarrow\text{ 2H}_{2(g)}\text{ + O}_{2(g)}[/tex]Step 2; Find the number of moles of oxygen atom
Recall, that 1 moles of a gas at STP is equivalent to 22.4 L/mol
Let x represents the number of moles of oxygen
[tex]\begin{gathered} \text{ 1 mole O}_2\text{ }\rightarrow\text{ 22.4l/mol} \\ \text{ x mole O}_2\text{ }\rightarrow\text{ 40.9 L} \\ \text{ Cross multiply} \\ \text{ 1 mole O}_2\text{ }\times\text{ 40.9L = x mole O}_2\times\text{ 22.4 L/mol} \\ \text{ Isolate x mole O}_2 \\ \text{ x mole O}_2\text{ = }\frac{1\text{ mole O}_2\times40.9\cancel{L}}{22.4\text{ }\frac{\cancel{L}}{mol}} \\ \\ \text{ x mole O}_2\text{ = }\frac{1\text{ }\times\text{ 40.9}}{22.4} \\ \text{ x mole O}_2\text{ = }\frac{40.9}{22.4} \\ \text{ x mole O}_2\text{ = 1.83 moles} \end{gathered}[/tex]The moles of O2 is 1.83 moles
Step 3; Find the number of moles of H2O using a stoichiometry ratio
Let x represents the number of moles of H2O
In the given reaction, 2 moles of water give 1 mole of O2
[tex]\begin{gathered} \text{ 1 mole O}_2\text{ }\rightarrow\text{ 2 moles H}_2O \\ \text{ 1.83 moles O}_2\text{ }\rightarrow\text{ x moles H}_2O \\ \text{ cross multiply} \\ \text{ 1 mole O}_2\text{ }\times\text{ x moles H}_2O\text{ = 2 moles H}_2O\text{ }\times\text{ 1.83 moles O}_2 \\ \text{ Isolate x moles H}_2O\text{ } \\ \text{ x moles H}_2O\text{ = }\frac{\text{ 2 moles H}_2O\times1.83moles\cancel{O_2}}{1mole\cancel{O_2}} \\ \\ \text{ x moles H}_2O\text{ = 2 }\times\text{ 1.83} \\ \text{ x moles H}_2O\text{ = 3.66 moles} \end{gathered}[/tex]The number of moles of H2O is 3.66 moles
Step 4; Find the mass of H2O using the formula below
[tex]\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }\times\text{ molar mass} \end{gathered}[/tex]Recall, that the molar mass of water is 18.0 g/mol
[tex]\begin{gathered} \text{ mass = 3.66 }\times\text{ 18} \\ \text{ mass = 65.88 grams} \end{gathered}[/tex]Therefore, the mass of H2O that reacted is 65.88 grams