蒟蒻才初二！！！
by chen_z @ 2022-10-18 13:42:18
tj里的
```
CxHyOz+(x+y/4-z/2)O2-> xCO2＋(y/2)H2O
```
看不懂，能否解释一下？
by chen_z @ 2022-10-18 13:44:28
$\mathsf{xC+2xO=\!=xCO_2}$
$\mathsf{yH+\frac{y}{2}O=\!=\frac{y}{2}H_2O}$
$\mathsf{xC+yH+(2x+\frac{y}{2})O=\!=xCO_2+\frac{y}{2}H_2O}$
$\mathsf{xC+yH+zO+(2x+\frac{y}{2}-z)O=\!=xCO_2+\frac{y}{2}H_2O}$
$\mathsf{C_xH_yO_z+(x+\frac{y}{4}-\frac{z}{2})O_2=\!=xCO_2+\frac{y}{2}H_2O}$
by 2017gdgzoi999 @ 2022-10-18 14:13:51
@[2017gdgzoi999](/user/120250) 谢谢！！！ ORZ orz （不过还是不会。。。
by chen_z @ 2022-10-26 13:41:42
$C_xH_yO_z=xC+yH+zO$
$=xCO_2-xO_2+\frac{y}{2}H_2O-\frac{y}{2}O+zO$
$=xCO2+\frac{y}{2}H_2O-(x\cdot2+\frac{y}{2}-z)O$
$\therefore C_xH_yO_z+(2x-\frac{y}{2}-z)O=xCO_2+\frac{y}{2}H_2O$
$\therefore C_xH_yO_z+(x-\frac{y}{4}-\frac{z}{2})O_2=xCO_2+\frac{y}{2}H_2O$
by cleverxia @ 2022-12-02 20:49:21