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REY [17]
3 years ago
6

Two burettes are set up. The first contains 0.15M NaOH and the second contains an HCl solution of unknown concentration. HCl is

dispensed into the reaction flask and Phenolphthalein is added as an indicator. At the end of the titration, 22.2ml of NaOH and 9.45ml HCl were used.
Required:
A. Write a balanced equation for the Acid-Base reaction.
B. What is the total volume of NaOH used in the titration?
C. How many moles of NaOH was used?
D. What is the total volume of HCl used in the titration?
E. What is the final concentration of the HCl solution?
Chemistry
1 answer:
damaskus [11]3 years ago
4 0

Answer:

A: NaOH + HCl = NaCl +H2O

B: Volume of NaOH = 22.2ml

C: mole=3.33\times10^{-3}

D: Volume of HCl = 9.45 ml

E: molarity=0.35 mole/litre = final concentration of HCl

Explanation:

Part A: Balanced chemical equaion

NaOH + HCl = NaCl +H2O

part B:

Volume of NaOH used in titration is 22.2 ml because volume is taken upto the end point of the titration.

Part C:

Calculation of moles of NaOH used:

mole=volume in litre\times molarity

mole=\frac{22.2}{1000} \times0.15

mole=3.33\times10^{-3}

Part D:

Volume of HCl used in titration is 9.45 ml because volume is taken upto the end point of the titration.

Part E:

N_1V_1=N_2V_2

0.15\times22.2=N_2\times 9.45

N_2=0.35

Normality =molarity \times n-factor

But in case of acid it is known as basicity and in case of base it is known as acidity.

in case of NaOH and HCl n-factor is 1;

hence

Normality=molarity;

molarity=0.35 mole/litre = final concentration of HCl

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8 0
4 years ago
A 3.3 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.3 g. The two substances react, rel
Zanzabum

Answer:

1.73g of CO2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass of CH3COOH from the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.

Next, we shall determine the limiting reactant of the reaction. This is illustrated below:

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.

From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.

Finally, can determine the mass of CO2 produced during the reaction.

In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:

From the balanced equation above,

84g of NaHCO3 reacted to produce 44g of CO2.

Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.

Therefore, 1.73g of CO2 is released during the reaction.

7 0
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Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon d
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Answer:1.

1.Balanced equation

C4H10 + 9 02 ==> 5H20 +4CO2

2. Volume of CO2 =596L

Explanation:

1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;

CxHy +( x+y/4) O2 ==> y/2 02 + xCO2

Where x and y are number of carbon and hydrogen atoms respectively.

For butane (C4H10)

x=4 and y=10

Therefore

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A.Compounds.
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