Calculam masa de potasiu ce se arde: m = 25/100x78
m= 19,5g K
4x39 2x94
4K + O₂→ 2 K₂O MK₂O= 2x39 +16
19,5g x = 94
x =19,5x2x94/ 4x39 x=23,5 g K₂O
Scriem ecutia reactiei chimice dintre oxidul de potasiu si apa:
94 2x56
K₂O + H₂O → 2 KOH MKOH = 39+16 +1
23,5g y = 56
y = 23,5x2x56/94 y = 28g KOH