Consider the reaction shown below. 2 H2(g) + O2(g) 2 H2O(g) (a) If 12.52 g of O2
ID: 1003323 • Letter: C
Question
Consider the reaction shown below. 2 H2(g) + O2(g) 2 H2O(g) (a) If 12.52 g of O2 react at 0.884 atm and 127°C, what volume of oxygen was used? WebAssign will check your answer for the correct number of significant figures. L (b) How many moles of water are produced in part a? WebAssign will check your answer for the correct number of significant figures. mol (c) If the water is produced at 760 mm Hg and 101°C, what volume of water was collected? WebAssign will check your answer for the correct number of significant figures. L
Explanation / Answer
Reaction is 2H2(g) + O2(g) = 2H2O(g)
a) If 12.52 g of O2 react at 0.884 atm and 127°C, what volume of oxygen was used?
We have,
Weight of oxygen = 12.52 g
Pressure P = 0.884 atm
Temperature T = 127°C = 400.15 K
R = 0.0820
Volume V = ?
We will be calculating volume by formula PV = nRT
But we don’t have moles of oxygen, let us calculate that by weight and molar mass of oxygen
n of O2 = 12.52/31.99 = 0.3913
V = nRT/P = 0.3913 * 0.0820 * 400.15 /0.884 = 14.52 L
b) How many number of moles are produced in part a)
As we can see in reaction two moles of water per mole of oxygen was produced.
As O2 is in less concentration, O2 is limiting reagent
0.391 moles of O2 requires 0.391 * 2 = 0.782 moles of H2O
c) Volume of water collected = ?
We have P = 760 mm Hg = 1.000 atm
n of water = 0.782 moles
R = 0.0820
T = 101°C = 374.15 K
V = nRT/P = 0.782 * 0.0820 * 374.15 / 1 = 23.99 L