Odpowiedź :
2Na + 2H2O --> 2NaOH + H2
mmol(H2O) = 16g/mol + 2*1g/mol = 18g/mol
Vmol(H2O) = 22,4 dm3/mol
2mol * 18g/mol = 36g
36g -- 22,4 dm3
30g - X
[tex]x = \frac{30g*22,4dm^{3} }{36g} =~18,67 dm3[/tex]
Odp: powstanie 18,67dm3 wodoru
2Na + 2H2O --> 2NaOH + H2
mmol(H2O) = 16g/mol + 2*1g/mol = 18g/mol
Vmol(H2O) = 22,4 dm3/mol
2mol * 18g/mol = 36g
36g -- 22,4 dm3
30g - X
[tex]x = \frac{30g*22,4dm^{3} }{36g} =~18,67 dm3[/tex]
Odp: powstanie 18,67dm3 wodoru