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nataly862011 [7]
2 years ago
11

A barium hydroxide solution is prepared by dissolving 2.06 g of Ba(OH)2 in water to make 32.9 mL of solution. What is the concen

tration of the solution in units of molarity? concentration: 0.364 The barium hydroxide solution is used to titrate a perchloric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between barium hydroxide and perchloric acid. chemical equation: 2HCIO4(a Ba(OH2(a)Ba CIO42H2o If 15.9 mL of the barium hydroxide solution was needed to neutralize a 8.50 mL aliquot of the perchloric acid solution, what is the concentration of the acid?
Chemistry
1 answer:
navik [9.2K]2 years ago
4 0

Answer:  1.36 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

To calculate the moles, we use the equation:

moles of solute= \frac{\text {given mass}}{\text {molar mass}}=\frac{2.06g}{171g/mol}=0.0120moles

Molarity=\frac{0.0120\times 1000}{32.9}=0.364M

The balanced reaction between barium hydroxide and perchloric acid:

2HCIO_4+Ba(OH_)2\rightarrow BaCIO_4+2H_2O

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HClO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ba(OH)_2

We are given:

n_1=1\\M_1=?\\V_1=8.50mL\\n_2=2\\M_2=0.364M\\V_2=15.9mL

Putting values in above equation, we get:

1\times M_1\times 8.50=2\times 0.364\times 15.9\\\\M_1=1.36M

Thus the concentration of the acid is 1.36 M

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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

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It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high
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Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

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The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)
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Answer:

a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

  • Na:  23
  • H: 1
  • C: 12
  • O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

  • NaHCO_{3} :23+1+12+16*3=84 g/mol
  • H_{3} C_{6} HO_{7} :1*3+12*6+1*5+16*7= 192 g/mol
  • CO_{2} :12+16*2= 44 g/mol
  • H_{2} O :1*2+16= 18 g/mol
  • Na_{3} C_{6} H_{5} O_{7} : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

grams of citric acid=\frac{1.4 g * 192 g}{252 g}

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

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