Answer :
Make the reaction: [tex]CH_4+2O_2=CO_2+2H_2O[/tex]
For 2*32=64 grams of oxygen, you can burn 2*16=32 grams of methane (that is, half).
So, our answer is .5g.
Other intermediate calculations are mollecular masses:
[tex]\mu_{CH_4}=A_C+4A_H=12+4=16 \\ mu_{O_2}=2A_O=2*16=32[/tex]
For 2*32=64 grams of oxygen, you can burn 2*16=32 grams of methane (that is, half).
So, our answer is .5g.
Other intermediate calculations are mollecular masses:
[tex]\mu_{CH_4}=A_C+4A_H=12+4=16 \\ mu_{O_2}=2A_O=2*16=32[/tex]
Answer:
0.28 g of [tex]CH_4[/tex]
Explanation:
You need a balanced equation first. [tex]CH_4 + 2O_2[/tex] ⇒ [tex]CO_2 + 2H_2O[/tex]
You need to find the number of moles of oxygen, which is mass divided by the Mr: 1 ÷ 32 = 1/32
Then you find the moles of methane using the mole ration oxygen to methane 2:1. 1/32 ÷ 2 = 1/64
Then you find the mass by multiplying the Mr with the number of moles:
1/64 × 18 = 0.28125 g = 0.28 g
