All categories

2 years ago

How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution

Chemistry

1 answer:

Lelechka [254]2 years ago

6 0

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
Volume of acid, HCl (Va) = 36 mL
Molarity of acid, HCl (Ma) = 1.63 M
Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

brainly.com/question/14356286

#SPJ1

You might be interested in

Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

3 0

3 years ago

With respect to NO and For the reaction NO + O 3 NO 2 + O 2 , the reaction was expressing concentrations as molarityare the unit

Sliva [168]

Answer:

CXbxcbfvbdffdffzsarg

Explanation:

6 0

3 years ago

How many orbitals are in the 5th energy level?

Alexandra [31]

Answer:7

Explanation:

8 0

3 years ago

Read 2 more answers

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

7 0

3 years ago

4.4 grams of CO2 represents how many moles?

jeka57 [31]

Answer:

1) Moles of CO 2 = Given mass / Molecular mass of CO 2 = 4.4 / 44 = 0.1 mole. 2) Molecules of CO2 in 0.1 moles of CO2 = 0.1 x 6.023 x 10 23 = 6.023 x 10 22 molecules 3) 44 gram (molecular wt of CO 2) contains 2 moles atom of oxygen therefore 4.4 gram of CO2 will contain = 2 /44 *4.4 = 0.2 moles atom of Oxygen.

8 0

2 years ago