Answer :
ANSWER
The mass of water formed is 28.52 grams
EXPLANATION
Given information
The volume of oxygen = 4.2L
The pressure = 5.0 atm
The temperature = 323.1K
To find the mass of H2O produced, follow the steps below
Step 1: Find the mole of oxygen using the ideal gas equation
[tex]\text{ PV = nRT}[/tex]Where
P = 5.0 atm
T = 323.1K
V = 4.2L
Step 2: Substitute the given data into the formula in step 1
[tex]\begin{gathered} \text{ Recall, that R is the unversal gas constant and it is 0.08205 L atm mol}^{-1}K^{-1} \\ \text{ PV = nRT} \\ \text{ 5}\times4.2\text{ = n }\times0.08205\times323.1 \\ 21\text{ = n }\times\text{ 26.510} \\ \text{ Divide both sides by 26.150} \\ \frac{21}{26.510}\text{ = n} \\ \text{ n = 0.792 mole} \end{gathered}[/tex]From the calculations above, the number of moles of oxygen is 0.792 mole
Step 3: Find the number of moles using a stoichiometry ratio
From the reaction above, 1 mole of oxygen will produce 2 moles of water
Let x represents the number of moles of water
[tex]\begin{gathered} \text{ x = 2 }\times\text{ 0.792} \\ \text{ x = 1.584 moles} \end{gathered}[/tex]The moles of water is 1.584 moles
Step 4: Find the mass of water using the below formula
[tex]\text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of water is 18.0 g/mol
[tex]\begin{gathered} \text{ 1.584 = }\frac{\text{ mass}}{\text{ 18}} \\ \text{ Mass = 1.584 }\times\text{ 18} \\ \text{ Mass = 28.52 grams} \end{gathered}[/tex]Hence, the mass of water formed is 28.52 grams