Answer:
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.
