Question

What mass of water is produced from the complete combustion of 4.70×10−3 g of methane?

Answer 1

Since the equation is balanced we can use the molar ratios to jump between amounts of substances. 
CH4 + 2O2 ---> CO2 + 2 H2O 

1.You can use a molar ratio to convert between mols methane to mols CO2: 
4.70×10−3 mol CH4 X (1 CO2 / 1 CH4) = 4.7x10^-3 mol CO2 X (44g/1mol) = .2068g CO2

2. You can do the same technique with the water. 
4.70×10−3 mol CH4 X (2 H2O / 1 CH4) = .0094 mol H2O X (18g/1mol) = .1692g H2O 

3. You can do the same technique with the O2. 
4.70×10−3 mol CH4 X (2 O2 / 1 CH4) = .0094 mol O2 X (32g/1mol) = .3008g O2

Answer 2

Answer:

$m_{H_2O}=1.06x10^{-2}gH_2O$

Explanation:

Hello,

In this case, when methane undergoes combustion, the reaction is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Thus, by stoichiometry, the mass of yielded water is:

$m_{H_2O}=4.70x10^{-3}gCH_4*\frac{1molCH_4}{16gCH_4}*\frac{2molH_2O}{1molCH_4}*\frac{18gH_2O}{1molH_2O}\\m_{H_2O}=1.06x10^{-2}gH_2O$

Best regards.