What mass of carbon dioxide is produced from the complete combustion of 8.00×10−3 g of methane? express your answer with the appropriate units.

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

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"