Consider the chemical reaction:
2H2O(l)→2H2(g)+O2(g)
How many moles of H2O are required to form 1.6 L of O2 at a temperature of 324 K and a pressure of 0.984 atm ?
Express your answer using two significant figures.

Respuesta :

Answer:

The answer to your question is: 0.118 moles of water

Explanation:

Data

moles of water = ?

Vol of O2= 1.6

Temp = 324°K

Pressure = 0.984 atm

R = 0.082 atm l / mol°K

Process

1.- Calculate the moles of O₂ using the ideal gas formula.

                                PV = nRT

                                [tex]n = \frac{PV}{RT}[/tex]

                                [tex]n = \frac{(1.6)(0.984)}{(0.082)(324)}[/tex]

                                [tex]n = \frac{1.5744}{26.568}[/tex]

                                       n = 0.059 moles of oxygen

2.- Find the moles of water

                                2 moles of water ------------------ 1 mol of oxygen

                                x                           ------------------   0.059 moles

                                x = (0.059 x 2) / 1

                                x = 0.118 moles of water

The number of moles of H2O required to form 1.6 L of O2 at a temperature of 324 K and a pressure of 0.984atm is 0.118moles

  • To calculate the number of moles of oxygen involved in the reaction can be calculated using the following expression:

PV = nRT

Where;

  1. P = pressure (atm)
  2. V = volume (L)
  3. n = number of moles (mol)
  4. R = gas law constant (0.0821 Latm/molK)
  5. T = temperature (K)

  • n = (0.984 × 1.6) ÷ (0.0821 × 324)

  • n = 1.5744 ÷ 26.6

  • n = 0.059mol of O2.

  • According to this reaction; 2H2O(l) → 2H2(g) + O2(g)

  1. 2 moles of H2O produce 1 mole of O2
  2. 0.059 moles of O2 will be produced when 0.059 × 2 = 0.118moles of H2O.

  • Therefore, the number of moles of H2O required to form 1.6 L of O2 at a temperature of 324K and a pressure of 0.984atm is 0.118moles.

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