Question

2 H2(g) + O2(g) → 2 H2O(g)

How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP?


a

56 L


b

77 L


c

35 L


d

67 L

Answer 1

77 L of water can be made.

Step-by-step explanation:

Molar mass of
`O_(2)` = 32 g/mol

So, 55 g of
`O_(2)` =
`(55)/(32)` mol of
`O_(2)` = 1.72 mol of
`O_(2)`

As hydrogen is present in excess amount therefore
`O_(2)` is the limiting reagent.

According to balanced equation, 1 mol of
`O_(2)` produces 2 mol of
`H_(2)O`.

So, 1.72 mol of
`O_(2)` produce
`(2* 1.72)` mol of
`H_(2)O` or 3.44 mol of
`H_(2)O`.

Let's assume
`H_(2)O` gas behaves ideally at STP.

Then,
`P_{H_(2)O}.V_{H_(2)O}=n_{H_(2)O}.R.T` , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here,
`n_{H_(2)O}` = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So,
`(1atm)* V_{H_(2)O}=(3.44mol)* (0.0821L.atm.mol^(-1).K^(-1))* (273K)`

`\Rightarrow`
`V_{H_(2)O}=77L`

Option (b) is correct.