calculate the mass and volume pf oxygen required at STP to convert 2.4 kg of graphite into carbon dioxide.
Answers
Answer:
The equation for the above reaction is as follows:
(graphite is an allotrope of carbon--C)
C + O2 -------> CO2
(s) (g) (g)
Therefore 1 mole of graphite requires 1 Mole of O2 to give 1 mole of CO2
Calculate the moles of graphite:
molar mass of C = 12
mass = 2400 g
moles = mass/molar mass
= 2400/12
=200 moles
if 1 mole of C requires 1 mole of O2
then 200 moles requires = 1 x 200 / 1
= 200 moles
Find mass of oxygen: (molar mass of O2= 32, moles=200)
mass = moles x molar mass
= 200 x 32
= 6400g
= 6.4kg of oxygen
Find the volume of oxygen:
Ideal gas law..... states that 1 mole of an ideal gas occupies 22.4 litres.
therefore 200 moles of O2 occupies= (22.4 x 200) liters
=4480 liters
volume of oxygen = 4480 litter
Answer:
at stp volume of 1 mole 22.6l and at ntp volume is 22.4 l
graphite is most stable thermodynic form of carbon
so formation of co2 from graphite take place according to reaction
c(s) +o2 give co2
where o2 and co2 are gas
2.4 kg graphite contain mole
= 2.4÷12 = 200 moles
apply unitry method
which says that
agar 1 rupees me 2 toffe ati h to
2 me kitni ayegi
1 mole 22.6
200 mole 200×22.6= 4520l
1 mole or 12 g of carbon require 32 g of oxygen to convert it to co2
200 mole or 200×12 g of carbon require 32×200×12 ÷12= 6400 g
12g.......32g
1g.........32÷12
200×12g........32×200×12÷12=6400g
hope it helps
bye