Question

Q.2 How many litres of oxygen will be produced by
decomposing 112L of ozone at STP?​

Answer 1

Answer:

10.5g⋅1 mole water÷18.02g=0.577 moles H2O

Now take a look at the balanced chemical equation

2H2(g)+O2(g)→2H2O(g)

Notice the 1:2 mole ratio that exists between oxygen and water. This means that 1 mole of oxygen will produce 2 moles of water.

Calculate the number of moles of oxygen by

0.577molesH2O .1 mole .O2÷2molesH2O

=

0.2885 moles

0.577molesH2O⋅1 mole O22molesH2O

=0.2885 molesO2

So, if 1 mole occupies 22.7 L at STP, then

0.2885 moles . ⋅22.7 L÷ 1mole=6.548 L

0.2885moles⋅22.7 L1mole=6.548 L

Rounded to three sig figs, the answer will be

VO2=6.55 L

SIDE NOTE More often than not, you will be required to use the old definition of STP, which implies a pressure of 1 atm and a temperature of 273.15 K. Under these conditions, 1 mole of any ideal gas occupies 22.4 L.

The current definition of STP implies a molar volume of 22.7 L, but if your instructor or teacher wants you to use the old value, simply redo the final calculation using 22.4 instead of 22.7 L.