Question

if 4g methane burns in 10g oxygen. 1) what is the amount of co2 formed 2. identfy the limiting reagent 3. calculate the amount of reactant left unreacted?
EQ: CH4 + 202 - CO2 + 2H20

Answer 1

CH4 + 202 ------->CO2 + 2H20
16g.m. 64gm. 44gm. 36gm
16gm of CH4 require = 64 gm O2
4 gm will CH4 require = (64/16)×4=
16gm O2
but we have only 10gm of O2
so O2 is limiting reagent
64 gm O2 produce = 44gm CO2
10 gm O2 will produce =(44/64)×10=6.9gm CO2
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64 gm O2 require =16gm CH4
10gm will require= (16/64)×10=2.5 gm CH4
Amount of CH4 remain unreacted=10-2.5=7.5 gm
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Amount of CO2 produce=6.9 gm
Limiting reagent is oxygen
Amount of CH4 remain unreacted =7.5gm