Answer:
22×10−3 g
Explanation:
Combustion equation of methane:
CH4 + 2O2 → CO2 + 2H2O
Given data:
mass of methane= 8.00 ×10−3 g
mass of carbon dioxide= ?
Solution 1:
m (CH4) = 8.00 ×10−3 g
v=m /M
v(CH4) = m (CH4)/ M (CH4)
v(CH4)= 8.00 ×10−3 g / 16 g/ mol
v(CH4)= 0.5 ×10−3 mol
according to balance chemical equation:
v(CH4) : v(CO2) = 1:1
v(CH4) = v(CO2)
M(CO2) = 44 g/mol
m (CO2) = M(CO2) × v(CO2)
m (CO2) = 44 g/mol ×0.5 ×10−3 mol = 22×10−3 g
Second method:
molecular weight of methane = 16 g/mol
molecular weight of carbon dioxide = 44 g/mol
mass of methane= 8.00×10−3 g
mass of carbon dioxide= ?
Solution:
mass of carbon dioxide= mass of methane × molecular weight of carbon dioxide / molecular weight of methane
mass of carbon dioxide= 8.00×10−3 g× 44 g/mol / 16 g/mol
mass of carbon dioxide= 22 × 10−3 g