Urgent!!! Topics of chemical difference method in the third year of junior high school

With methane aml, then oxygen will have (30-a) ml if methane is depleted.

ch4+2o2=co2+2h2o

a 2a a △v=2a

As can be obtained from the question, there is a change of 14ml. So.

a/（30-a）=7/23

If the oxygen is depleted.

ch4+ 2o2= co2 +2h2o

15-a 2 30-a 15-a 2 v = 30-a so a = 14

a/（30-a）=7/8

That's a bit out of place.

ch4+o2=2h2o+co2

2 volumes of gas are reacted to produce 1 volume of gas.

There is a volume reduction of 30-16 = 14 ml.

It is the volume of H20 that is generated, and CO2 is produced 7ml

Reactions CH47O27

There are 16-7 = 9ml of gas left.

Divide the situation. 02 overdose, i.e., 9ml, is 02 ch4:o2=7:(7+9)ch4 overdose ch4:o2=(7+9):7

ch4+2o2=co2+2h2o

x y 14

x=7 y=14

30ml reflects 21ml and 9ml remaining

So the volume ratio might be (7+9):14=8:7 or 7:(14+9)=7:23

Let methane a ml, then oxygen (30-a) ml

The first case: excess methane. In the calculation of the chemical equation, the oxygen consumption is the entire oxygen content in 30ml, and the methane is not involved in the reaction in addition to the reaction part, so it should be calculated according to the amount of oxygen Note: Water is liquid at room temperature.

ch4 + 2 o2 = co2 + 2 h2o δva 2a a 2a

30-16=14ml

As can be seen from the question, the multiple of oxygen is the same as that of δv.

v o2=14ml

vch4=30-14=16ml

Volume ratio 8 7

The second: excess oxygen, calculated as methane.

ch4 + 2 o2 = co2 + 2 h2o δva 2a a 2a

30-16=14ml

As can be seen from the question, methane and δV are 1 2 relationship, VCH4 = 14 2 = 7MLVo2 = 30-7 = 23 ml

Volume ratio 7 23

1) A:. 2) Solution: Na2CO3 2HCl 2NaCl CO2 Knowing the Pin Before Fierce H2O Because M(CO2), so 44 106 m(CO2) M(Na2CO3) Solution M(Na2CO3) mass fraction is hopeful, Xie Huixiang Xie.