what volume of air(containing 20% Oxygen by volume) will be required to burn completely 10cm³ each of methane and acetylene?
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Methane is CH4
Acetylene is C2H2
Rxn for CH4
CH4 + 2(O2 ) — (CO2) +2(H2O)
1cm^3 of CH4 will react with 2cm^3 of O2
Thus 10cm^3 with (2×10) cm^3
=20cm^3
Rxn for C2H2
2(C2H2) + 5(O2) — 4(CO2) + 2(H2O)
Here 2 with 5
So 1 with 5/2
And 10 with 2.5×10=25
So total O2 required is 20+25=45
In 100 cm^3 there is 20cm^3 of O2
So 20 in 100
1 in 100/20=5
45 in 5×45=225cm^3 of air is required
good afternoon_____________✌️✌️✌️✌️
_________________________________________________________
Methane is CH4
Acetylene is C2H2
Rxn for CH4
CH4 + 2(O2 ) — (CO2) +2(H2O)
1cm^3 of CH4 will react with 2cm^3 of O2
Thus 10cm^3 with (2×10) cm^3
=20cm^3
Rxn for C2H2
2(C2H2) + 5(O2) — 4(CO2) + 2(H2O)
Here 2 with 5
So 1 with 5/2
And 10 with 2.5×10=25
So total O2 required is 20+25=45
In 100 cm^3 there is 20cm^3 of O2
So 20 in 100
1 in 100/20=5
45 in 5×45=225cm^3 of air is required
Rimjhim58:
in the question it is 10cm³ of methane and acetylene
not ml
how can we write ml instead of cm³
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