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

`m_(H_2O)=1.06x10^(-2)gH_2O`

Step-by-step 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*(1molCH_4)/(16gCH_4)*(2molH_2O)/(1molCH_4)*(18gH_2O)/(1molH_2O)\\m_(H_2O)=1.06x10^(-2)gH_2O`

Best regards.