Consider the following chemical reaction: 2H2O (l) →2H2 (g) + O2 (g) What mass of H2O is required to form 1.3 L of O2 at a temperature of 325 K and a pressure of 0.991 atm? Express your answer using two significant figures.

1.73g of H2O

Explanation:

The following data were obtained from the question:

Volume (V) of O2 = 1.3L

Temperature (T) = 325 K

Pressure (P) = 0.991 atm

Gas constant (R) = 0.0821 atm. L/Kmol

Next, we shall determine the number of mole (n) of O2 produced from the reaction. This can be obtained by using the ideal gas equation as shown below:

PV = nRT

0.991 x 1.3 = n x 0.0821 x 325

Divide both side by 0.0821 x 325

n = (0.991 x 1.3) / (0.0821 x 325)

n = 0.048 mole

Therefore, 0.048 mole of O2 was produced from the reaction.

Next, we shall determine the number of mole H2O that produce 0.048 mole of O2. This is illustrated below:

2H2O (l) → 2H2 (g) + O2 (g)

From the balanced equation above,

2 moles of H2O produced 1 mole of O2.

Therefore, Xmol of H2O will produce 0.048 mole of O2 i. e

Xmol of H2O = (2 x 0.048) / 1

Xmol of H2O = 0.096 mole

Therefore, 0.096 mole of H2O was used in the reaction.

Finally, we shall convert 0.096 mole of H2O to grams. This is illustrated below:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Number of mole H2O = 0.096 mole

Mass of H2O = ... ?

Mole = mass / molar mass

0.096 = mass / 18

Cross multiply

Mass = 0.096 x 18

Mass of H2O = 1.73g

Therefore, 1.73g of H2O is required for the reaction.