Question

4.2 L of O2 reacts with excess H2 to form H20 according to the balanced equationbelow:2H2(g) + O2(g) → 2H2O(g)If the pressure is 5.0 atm and the temperature is 323.1 K, how many grams of H20are formed?

Answer 1

The mass of water formed is 28.52 grams

Step-by-step 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

`\text{ PV = nRT}`

Where

P = 5.0 atm

T = 323.1K

V = 4.2L

Step 2: Substitute the given data into the formula in step 1

`\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}*4.2\text{ = n }*0.08205*323.1 \\ 21\text{ = n }*\text{ 26.510} \\ \text{ Divide both sides by 26.150} \\ (21)/(26.510)\text{ = n} \\ \text{ n = 0.792 mole} \end{gathered}`

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

`\begin{gathered} \text{ x = 2 }*\text{ 0.792} \\ \text{ x = 1.584 moles} \end{gathered}`

The moles of water is 1.584 moles

Step 4: Find the mass of water using the below formula

`\text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of water is 18.0 g/mol

`\begin{gathered} \text{ 1.584 = }\frac{\text{ mass}}{\text{ 18}} \\ \text{ Mass = 1.584 }*\text{ 18} \\ \text{ Mass = 28.52 grams} \end{gathered}`

Hence, the mass of water formed is 28.52 grams