What mass of oxygen is needed for the complete combustion of 6.00×10−3 g
g of methane?
CH4 + 2O2 → CO2 + 2H2O
M(CH4) = 16 g/mol
n(CH4) =6.00×10−316=0.375×10−3 mol= \frac{6.00 \times 10^{-3}}{16} = 0.375 \times 10^{-3} \; mol=166.00×10−3=0.375×10−3mol
According to the reaction:
n(O2) = 2n(CH4) =2×0.375×10−3=0.75×10−3 mol= 2 \times 0.375 \times 10^{-3} = 0.75 \times 10^{-3} \;mol=2×0.375×10−3=0.75×10−3mol
M(O2) = 32 g/mol
m(O2) =0.75×10−3×32=24×10−3=0.024 g= 0.75 \times 10^{-3} \times 32 = 24 \times 10^{-3} = 0.024 \;g=0.75×10−3×32=24×10−3=0.024g
Answer: 0.024 g
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