Question

Se trateaza o roca de calcar cu 700g solutie H2SO4 de concentratie 40%. Calculati masa de roca de CaCO3, volumul de gaz rezultat si compozitia procentuala a gazului.Multumesc.

Answer 1

CaCO₃+H₂SO₄→CaSO₄+H₂O+CO₂
c=md*100/ms
md=c*ms/100=700*40/100=280 g H₂SO₄
n=m/M=280/98=2.85 moli H₂SO₄
1 mol CaCO₃....................1 mol H₂SO₄
x.....................................2.85
x=2.85 moli CaCO₃
M CaCO₃=40+12+16*3=100 g/mol
n=m/M              m=n*M=2.85*100=285 g CaCO₃

1 mol H₂SO₄....................1 mol CO₂
2.85.................................y                            y=2.85 moli CO₂
1 mol ocupa 22.44 L
1 mol CO₂.....................................22.4 L
2.85..............................................z
z=22.4*2.85=63.84 L CO₂

M CO₂ = 12+32=44 g/mol
44 g CO₂....................12 g C.................32 g O₂
100............................m g C..................n g O₂
m=100*12/44=27.27% C
n=100*32/44=72.73% O₂