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
Answers
Answered by
2
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
---------------–----------------------------------
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
--------------–-----------------------------
Amount of CO2 produce=6.9 gm
Limiting reagent is oxygen
Amount of CH4 remain unreacted =7.5gm
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
---------------–----------------------------------
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
--------------–-----------------------------
Amount of CO2 produce=6.9 gm
Limiting reagent is oxygen
Amount of CH4 remain unreacted =7.5gm
error21307:
how many electrons will be present in 16g of ch4?
in Carbon there r 6 electron and in hydrogen there is 1 electron
so 1 molecule of CH4 will have electron
oh thankssss
6.022 * 10^24
16 gm Of CH4 contain 6.023×10²³ molecules .
6.023×10²³ will contain 6.023×10²⁴ electron
Similar questions