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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

Respuesta :

Answer:

77 L of water can be made.

Explanation:

Molar mass of [tex]O_{2}[/tex] = 32 g/mol

So, 55 g of [tex]O_{2}[/tex] = [tex]\frac{55}{32}[/tex] mol of  [tex]O_{2}[/tex] = 1.72 mol of  [tex]O_{2}[/tex]

As hydrogen is present in excess amount therefore  [tex]O_{2}[/tex] is the limiting reagent.

According to balanced equation, 1 mol of  [tex]O_{2}[/tex] produces 2 mol of [tex]H_{2}O[/tex].

So, 1.72 mol of [tex]O_{2}[/tex] produce [tex](2\times 1.72)[/tex] mol of [tex]H_{2}O[/tex] or 3.44 mol of [tex]H_{2}O[/tex].

Let's assume [tex]H_{2}O[/tex] gas behaves ideally at STP.

Then, [tex]P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T[/tex]   , 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, [tex]n_{H_{2}O}[/tex] = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, [tex](1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)[/tex]

  [tex]\Rightarrow[/tex]  [tex]V_{H_{2}O}=77L[/tex]

Option (b) is correct.

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