дано
m(AL(OH)3) = 1.56 g
m(HCL)-?
Al(OH)3 + 3HCL-->AlCL3+3H2O
M(AL(OH)3)= 78 g/mol
n(Al(OH)3) = m/M = 1.56 / 78 = 0.02 mol
n(Al(OH)3) = 3n(HCL)
n(HCL) = 0.02 * 3 = 0.06 mol
M(HCL) = 36.5 g/mol
m(HCL) = n*M = 0.06 * 36.5 = 2.19 g
ответ 2.19 г
дано
m(AL(OH)3) = 1.56 g
m(HCL)-?
Al(OH)3 + 3HCL-->AlCL3+3H2O
M(AL(OH)3)= 78 g/mol
n(Al(OH)3) = m/M = 1.56 / 78 = 0.02 mol
n(Al(OH)3) = 3n(HCL)
n(HCL) = 0.02 * 3 = 0.06 mol
M(HCL) = 36.5 g/mol
m(HCL) = n*M = 0.06 * 36.5 = 2.19 g
ответ 2.19 г