64g 2 * 18 g 98g
Ca (s) + 2 H2O (l) = Ca(OH)2 (aq) + H2 (g)
2g/y g xg / 20g zg
nCa= 64 g/mol
nH2O= 18 g/mol
nCa(OH)2 = 98 g/mol
x = 1,125 9 H2O < m H2O I => m H2O (exces) = 20 - 1,125 = 18,875 g
y = 35,55 g Ca > m Ca I
z = 98 * 2/ 64 = 3 g Ca(OH)2
m solutie finala = m H2O (exces) + m Ca(OH)2 = 3 + 18,875 = 21,875 g solutie
c final = md*100/ms= 3*100/21,875 = 13.71 %