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Masa (Na2CO3) (g) V gastado (HCl)

1 0.31 12.5 mL

2 0.3 12.5 mL

3 0.31 12.8 mL

PROMEDIO 0.306 12.6 mL

Masa (NaOH) V gastado (HCl)

1 2.02 23 mL

2 24 mL

PROMEDIO 2.02 23.5 mL

HCL

N= 〖m Na〗_(2〖CO〗_3 )/(( V gastado HCl )( weq.〖 Na〗_(2〖CO〗_3 )) ) Weq.= PM/(# eq.)=(106 g/mL)/(2 eq./mol)=53 g/eq.

N= (0.306 g)/(( 0.126 L )(53 g/eq.) )=0.458 eq./L

%Error=|(0.5-0.45)/0.5|x 100= 10%

NaOH

N(NaOH) = ((V.gastado HCl)(C HCl))/((V NaOH ) )

N(NaOH)= (0.0235 L)(0.45)/((0.02L ) )= =0.52

%Error=|(0.5-0.52)/0.5|x 100= 4%

CUESTIONARIO.

1.

2.

3.

4. M=n/v → n= MV= (0.5)(0.1L) = 0.05 mol

n=m/PM→ m = (n)(PM) = (0.05)(36.5) = 1.825 g

% m HCl = mHCl / mT x 100 -→ mT = mHCl / (%mHCl / 100) = 100 mHCl / %mHCl

(100)(1.825)

mT = = 4.87 gr

37.5

ρHCl = 1.17 ml / gr m= 4.87 g

ρ = m/V --- V = m/ρ = 4.87/1.17 = 4.16 ml

5.Determine la masa de NaOH que se requirió para preparar 10 mL desolución 0.5 M de NaOH.

M = 0.5 M

Vsolución = 100 mL = 0.1 L

PM HCl = 40 g/mol

M=n/L → n= (M)( L ) = (0.5)(0.1)= 0.05 mol

n=m/PM → m= n PM = (0.05)(40) = 2 g

6. Resuelto anteriormente

7. Resuelto anteriormente

8. a)

i.- 2 mL de H2SO4 15 N

N=(eq.gr.)/(L(soluto))

eq.gr =(N)(L)

Eq. - gr = 15 (2x10-3L) = 0.03

ii.- 50 mL de H2SO4 0.25 N

N=(eq.gr.)/(L(soluto))

eq.gr =(N)(L)

eq. - gr = (0.25)(0.050) = 0.0125

b)

N=(m(soluto))/(V(solucion)(PM/(#valencia)))

m(soluto) =(N)(V)( PM/(#valencia))

m soluto = (3) (0.5) (49) = 73.5 gr

%mH2SO4 = (mH2SO4 / mT) x 100

mT = 100 mH2SO4 / % m H2SO4 → mT = (100 x 73.5) / 93= 79.03 g

ρ = m/ V → V= m / ρ → V = 79.03 / 1.19 = 66.41 mL

c)

N=(m(soluto))/(V(solucion)(PM/(#valencia)))

m(soluto) =(N)(V)( PM/(#valencia))

msoluto = (0.002)(18)(36.5) = 1.314 gr

%mHCl = (mHCl / mT) x 100

mT = 100 mHCl / % m HCl

mT= (81.314 x 100) / 38 = 3.457 g

ρ = m/ V → V= m / ρ → V = 3.457 gr / 1.14 gr/ ml = 2.90 mL

d)

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