Question

determinati partea da masa a sarii obtinute in urma reactiei a 40cm cub de hidroxid de potasiu de6.2%(p=1.055g/cm cub ) cu 10 cm cub de acid fosforic (p=1.075g/cm cub) cu partea de masa 13.76%

Answer 1

ro1=ms1/V1=>ms1=ro1*V1=1.055*40=42.2g sol KOH

C=md/ms*100=>md=6.2*42.2/100=2.6164g KOH

mH2O=39.5836g

n=m/M=2.6164/56=0.04672 moli KOH

ro2=ms2/V2=>ms2=1.075*10=10.75g sol H3PO4

C=md/ms*100=>md=10.75*13.76/100=1.4792g H3PO4

mH2O=9.2708g

n=m/M=1.4792/98=0.015 moli H3PO4

KOH în exces -> 0.04672>0.015

3KOH + H3PO4 = K3PO4 + 3H2O

3*56g KOH....98g H3PO4......212g K3PO4......54g H2O
x=2.535g........1.4792g....y=3.2g....z=0.815g

m exces = 2.6164-2.535=0.0814g

md final = m exces + K3PO4=0.0814+3.2=3.2814g

mH2O final = 0.815+39.5836+9.2708= 49.6694g

ms final = md + mH2O=3.2814+49.6694 = 52,95g

C=m sare / msf *100=3.2/52.95*100=6%

Nu sunt sigur..