Answer:
1. 9.4 grams of methane produce 25.85 grams of CO2
2.Grams of water produced = 11.81 grams
3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams
4.Amount of methane consumed = 46.9 grams
5. Grams of Co2 produced = 8.32 grams
Explanation:
Molar masses :
Methane = CH4 = mass of C + 4x (mass of H)
CH4 = 12 +4(1) = 16 grams
1 mole of CH4 = 16 gram
Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)
Carbon Dioxide =CO2 = mass of C + 2(mass of O)
= 12 + 2(16)
= 44 grams (1 mole of CO2 = 44 gram )
Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)
1 mole of each molecule is equal to their molar masses
The balanced equation is :
[tex]1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)[/tex]
According to Stoichiometry :
1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O
1. From the equation ,
1 mole of methane produce =1 mole of CO2
16 gram of methane = 44 gram of CO2
1 gram of methane =
[tex]\frac{44}{16}[/tex] gram of CO2
9.4 gram of CH4 =
[tex]\frac{44}{16}\times 9.4[/tex] gram of CO2
= 25 .85 gram of CO2
2.
2 mole of O2 produces = 2 mole of H2O(water)
1 mole of O2 produces = 1 mole of H2O
32 gram of O2 = 18 gram of water
1 gram of O2 =
[tex]\frac{18}{32}[/tex]
21 gram of O2 =
[tex]\frac{18}{32}\times 21[/tex]
11.81 gram of water
3. 1 mole of CH4 = 2 mole of O2
16 gram of CH4 = 2(32) = 64 grams of O2
64 gram of O2 needs = 16 grams of CH4
1 gram of O2 needs =
[tex]\frac{16}{64}[/tex]
10.1 gram of O =
[tex]\frac{16}{64}\times 10.1[/tex] of CH4
= 2.52 gram
4.
1 mole of CO2 is produced from = 1 mole of CH4
44 gram of CO2 is produced from 16 gram of CH4
1 gram CO2 =
[tex]\frac{16}{44}[/tex] gram of CH4
129 gram of CO2 =
[tex]\frac{16}{44}\times 129[/tex] gram of CH4
= 46.90 grams
5.
2 mole of O2 produce = 1 mole of CO2
2x 32 gram of O2 = 44 gram of CO2
1 gram of O2 =
[tex]\frac{44}{64}[/tex] of CO2
12.1 gram of O2 produce=
[tex]\frac{44}{64}\times 12.1[/tex] of CO2
= 8.318 gram
Note : Write the quantity give on left side of "="
write the substance asked on right side of "="
