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IgorC [24]
2 years ago
10

A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

orms which is then filtered and dried. The mass of this precipitate is 0.284 g. The limiting reagent in the salt mixture was later determined to be CaCl2 ∙ 2 H2O.
a. Write the molecular form of the equation for the reaction.
b. Write the net ionic equation for the reaction
c. How many moles of CaCl2 ∙ 2 H2O reacted in the reaction mixture?
d. How many grams of CaCl2 ∙ 2 H2O reacted in the reaction mixture?
e. How many moles of K2C2O4 ∙ H2O reacted in the reaction mixture?
f. How many grams of K2C2O4 ∙ H2O reacted in the reaction mixture?
g. How many grams of K2C2O4 ∙ H2O in the salt mixture remain unreacted?
h. What is the percent by mass of each salt in the mixture?
Chemistry
1 answer:
Alecsey [184]2 years ago
6 0

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

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How many grams of CO are needed with an excess of Fe2O3 to produce 27.9 g Fe? Please show work. Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2
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Answer:

<u>20.98 gram of CO</u>

<u></u>

Explanation:

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We need to know which is limiting reagent to calculate the amount of product or vice-versa.

In this question , it is already given that Fe2O3  is excess reagent .Hence<u> CO must be the limiting reagent.</u>

The balanced equation is :

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So ,

1 gram of Fe is produced from =

\frac{84}{111.68} of CO

27.9 g of Fe is produced from=

\frac{84}{111.68}\times 27.9 gram of CO

<u>= 20.98 gram of CO</u>

= 21 gram of CO (In round figures)

<u />

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