7b 17. Potassium acid phthalate, KHC 8 H 4 O 4 , or KHP, is used in many laborat

Question

7b

17. Potassium acid phthalate, KHC8H4O4, or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed in air. A 0.3754 g sample of KHC8H4O4 reacts with 37.33 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH? KHC8H4O4(aq) + NaOH(aq) ? KNaC8H4O4(aq) + H2O(l). _____M

19. How many grams of solid calcium carbonate must decompose to produce solid calcium oxide and 540.0 mL carbon dioxide gas at 298°C and 727 torr? ______g

9. Milk of magnesia, an over-the-counter antacid liquid, is a suspension of magnesium hydroxide in water. It neutralizes the hydrochloric acid in the stomach to alleviate some stomach symptoms associated with excess acid. In a laboratory, 1.11 g Mg(OH)2 is added to 215.9 mL of 0.1790 M HCl in water. They react according to the following equation.

Mg(OH)2 + 2 HCl ? MgCl2 + 2 H2O

(a) What is the limiting reagent in this reaction?

(b) How much magnesium chloride is produced?

Explanation / Answer

17)

the Reaction is

KHC8H4O4(aq) + NaOH(aq) ? KNaC8H4O4(aq) + H2O(l)

moar mass of KHC8H4O4 = 204.44 gm/mole

that mean 1 mole of KHC8H4O4 = 204.44gm then

0.3574 gm = 0.3574/204.44

= 0.001748 mole of KHC8H4O4

According to reaction 1 mole of KHC8H4O4 react with 1 mole of NaOH then 0.001748 mole KHC8H4O4react with 0.001748 mole of NaOH

thus mole of NaOH = 0.001748 mole

37.33 ml = 0.03733 L

Molarity = no. of mole of solute/ volume of solution in liter

Molarity of NaOH = 0.001748/0.03733 = 0.047M

Molarity of NaOH = 0.047M

19)

We need a reaction:

Solid Calcium Carbonate = CaCO3

Solid Calcium Oxide = CaO

Apparently, there is loss of CO2...

the reaction must be the termal decomposition of CaCO3

this is given as

CaCO3(s) ---> CaO(s) + CO2(g)

note that all is solid but CO2 gas...

then

we are given V,T,P which are typically related to a gas

we can use the ideal gas law; so we cna relate to mol/mass

Apply law

PV = nRT

n = PV/(RT)

R is the ideal gas constant, use 0.082 for Liter, Kelvin and Atm

change to Liter; Kelvin and Atm instead of mL; Celcius and torr

T = 298C = 298 + 273 K = 571 K

298C is a very common number used, it is typically 273 + 25; that is, ambient Temperature when changed form Celcius to Kelvin.

Please consider that this may be Kelvin already and not C as stated...

P = 1 atm --> 760 torr

P = 727 torr --> 727/760 atm = 0.96 atm

For Volume

V = 1000 mL = 1 L

V = 540/1000 = 0.54 L

then

n = PV/(RT)

n = 0.96*0.540 / (0.082*571) = 0.0111 mol of CO2 gas

now, relate to stoichiometry

1 mol of CaCO3 --> 1 mol of CO2

then

0.0111 mol of CO2 --> 0.0111 mol of CaCO3

but we need mass so

1 mol of CaCO3 = 100.01 g/mol

0.0111 mol of CaCO3 = ?

mass = mol*MW = 0.0111*100.01 = 1.110 g of CaCO3 were used in order to get such amount of volume of CO2 in such conditions of T.P

mass of CaCO3 = 1.110 g

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