Question

exercițiul 19.13 si 19.14 va rog

Answer 1

19.13.a) MCuO=ACu+AO=64+16=80g/mol
m=160g CuO

n=m/M=160/80=2 moli CuO

b) MCO2=44g/mol. m=8.8g CO2

n=m/M=8.8/44=0.2 moli CO2

c) MH2SO4=98g/mol. m=9.8g H2SO4

n=m/M=9.8/98=0.1 moli H2SO4

d) MHCl=36.5g/mol. m=3.65g HCl

n=m/M=3.65/36.5=0.1 moli HCl

19.14. a) MFe=56g/mol. n=2 moli Fe

n=m/M=>m=n*M=2*56=112g Fe

b) MN2=28g/mol. n= 4 moli N2

n=m/M=>m=4*28=112g N2

c) MCl2=71g/mol. n=0.1 moli Cl2

m=0.1*71=7.1g Cl2

d) MH2SO4=98g/mol. n=5 moli H2SO4

m=5*98=490g H2SO4