Rozwiązanie:

$K_{2}S{O}_x$	$AlP{O}_x$	$Cr{O}_x$	$N_2{O}_x$	$Mg{\left(N{O}_x\right)}_2$
158 u	106 u	68 u	92 u	148 u
x = 3	x = 3	x = 1	x = 4	x = 3

 

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Wyjaśnienie:

• $K_{2}S{O}_x$  

$158u=2\cdot 39u+32u+x\cdot 16$  

$x=\frac{158u-2\cdot 39u-32u}{16}=3$  

 

• $AlP{O}_x$  

$106u=27u+31u+x\cdot 16$  

$x=\frac{106u-27u-31u}{16}=3$  

 

• `CrO_x` 

$68u=52u+x\cdot 16$  

$x=\frac{68u-52u}{16}=1$  

 

• $N_2{O}_x$  

$92u=2\cdot 14u+x\cdot 16$  

$x=\frac{92u-28u}{16}=4$  

 

• $Mg{\left(N{O}_x\right)}_2$  

$148u=24u+2\cdot 14u+x\cdot 16$  

$x=\frac{148u-24u-28u}{16}=3$