Question

Se amesteca 200 g solutie de HCL egal cu 14,6% cu 300 g solutie NH3 egal cu 10% in vederea obtinerii NH4CL . CALCULATI MASA DE SUBSTANTA IN EXCES SI CU CIT
NUMARUL DE MOLI DE CLORURA DE AMONIU FORMATA

Answer 1

C=md/ms*100 => md=C*ms/100=14,6*200/100=29,2g HCl

C=md/ms*100 => md=C*ms/100=10*300/100=30g NH3

MHCl=36,5g/mol

MNH3=17g/mol

MNH4Cl = AN+4AH+ACl = 14+4*1+35,5=53,5g/mol

n=m/M=29,2/36,5=0,8 moli HCl

n=m/M=30/17=1,7647 moli NH3

1,7647 > 0,8 => NH3(amoniac) in exces

HCl + NH3 = NH4Cl

36,5g HCl..............17g NH3........53,5g NH4Cl
29,2g HCl..........x=13,6g NH3.......y=42,8g NH4Cl (sare)

m exces => 30-13,6=16,4g NH3

36,5g HCl.........................1mol NH4Cl
29,2g HCl.........................x=0,8 moli NH4Cl