Review Assignment Chm 152 Reactions in Aqueous Solutions Name: 4) Reagent grade

Question

Review Assignment Chm 152 Reactions in Aqueous Solutions Name: 4) Reagent grade cond grade sulfuric acid, HSO, is between 95 and 98% acid in water. To precisely know the concentration, one on, one needs to do a titration of a sample of the acid. Aldrich gives the sulfurie acid as 1.840 g'ml. The molar mass of sulfuric acid is 98.08g/mol. sul density of a) Based on density and molecular mass, what is the estimated concentration of the st mass, what is the estimated concentration of the stock sulfuric acid? (mol/L) A student dilutes 100 mL of the stock sulfuric acid into 1.000 L distilled water. The student then uses a 10.0 mL sample of this dilution for a titration. It takes 0.9978 M NaOH solution to reach the equivalent point. i) Write the balanced molecular equation for this neutralization reaction; remember b) lation for titration.it takes 15.0 m n for a titration. It takes 35.70 mlL, of HzSOs has two protons that react with a strong base: ii) Write the complete ionic equation for this reaction: ii) Write the net ionic equation for this reaction: iv) What is the concentration of sulfuric acid in the dilution (mol/L)? v) What is the concentration of sulfurie acid in the stock solution? (molVL) Page 2 of 3

Explanation / Answer

4)

a)

for 1000 ml = 1 L

mass of H2S04 = 1.84 x 1000 = 1840

moles = mass / molar mass

so

moles of H2S04 = 1840 / 98.08 = 18.76

now

molarity = moles / volume (L)

so

molarity of H2S04 = 18.76 / 1 = 18.76

so

concentration of stock solution is 18.76 M

b)

i)

H2S04 + 2 NaOH ---> Na2S04 + 2H20

ii)

2H+ + S042- + 2Na+ + 2OH- ---> 2Na+ + S042- + 2H20

iii)

2H+ + 2OH- ---> 2H20 (l)

iv)

we know that

moles = molarity x volume (L)

so

moles of NaOH = 0.9978 x 35.7 x 10-3 = 35.62146 x 10-3

now

H2S04 + 2 NaOH ---> Na2S04 + 2H20

we can see that

moles of H2S04 = 0.5 x moles of NaOH

so

moles of H2S04 = 0.5 x 35.62146 x 10-3 = 17.81073 x 10-3

volume = 10 ml

so

conc = moles x 1000 / volume (ml)

conc = 17.81073 x 10-3 x 1000 / 10

conc = 1.781073

so

concentration of H2S04 in the dilution is 1.781073

v)

now

for dilution

M1V1 = M2V2

so

M1 x 100 = 1.781073 x 1000

M1 = 17.81073

so

the stock solution of H2S04 is 17.81073 M

3)

moles = molarity x volume (L)

so

moles of Na+ = 0.15 x 470 x 10-3 = 0.0705

b)

moles of Na+ = 0.249 x 30 x 10-3 = 0.00747

c)

total moles of Na+ = 0.0705 + 0.00747 = 0.07797

d)

total volume = 470 + 30 = 500 ml

e)

[Na+] = moles x 1000 / volume (ml)

so

[Na+] = 0.07797 x 1000 / 500

[Na+] = 0.15594 M

5)

a)

moles = mass / molar mass

so

moles of Zinc = 23.05 / 65.38

moles of zinc = 0.35255

b)

we can see that

moles of H2 = moles of Zn reacted = 0.35255

now

mass = moles x molar mass

so

mass of H2 = 0.35255 x 2 = 0.705

so

0.705 grams of H2 can be produced

c)

molar mass of HCl is 36.46 g / mol

d)

we know that

moles = molarity x volume (L)

so

moles of HCl = 1.2 x 500 x 10-3

moles of HCl = 0.6

e)

we can see that

moles of H2 produced = 0.5 x moles of HCl

so

moles of H2 = 0.5 x 0.6 = 0.3

mass of H2 = 0.3 x 0.2 = 0.6 grams

f)

we can see that

moles of HCl required = 2 x moles of Zn = 2 x 0.35255 = 0.705

but

only 0.6 moles of HCl is present

so

HCl is the limiting reagent

g)

so

mass of hydrogen produced is 0.6 grams

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