All categories

3 years ago

If 5.12 liters of a 2.75 M phosphoric acid is neutralized by magnesium hydroxide solution of 4.00 M.

Chemistry

2 answers:

Tresset [83]3 years ago

8 0

Answer:

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

Molarity of base (Mb) = 4M

Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 3

Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

Vb = 5.28 L

Therefore, the volume of the base is 5.28 L

worty [1.4K]3 years ago

6 0

Answer:

The volume of the base Mg(OH)2 used is 5.28 L

Explanation:

Step 1: Data given

Volume of phosphoric acid = 5.12 L

Concentration of phosphoric acid = 2.75 M

concentration of magnesium hydroxide = 4.00 M

Step 2: The balanced equation

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

Step 3: Calculate the volume of magnesium hydroxide needed

a*Cb*Vb = b*Ca*Va

⇒with a = the coefficient of H3PO4 = 2

⇒with Cb = the concentration of Mg(OH)2 = 4.00 M

⇒with Vb = the volume of Mg(OH)2 = TO BE DETERMINED

⇒with b = the coefficient of Mg(OH)2 = 3

⇒with Ca = the concentration of H3PO4 = 2.75 M

⇒with Va = the volume of H3PO4 = 5.12 L

2 * 4.00 * Vb = 3 * 2.75 * 5.12

Vb = (3*2.75 * 5.12) / (2*4.00)

Vb = 5.28 L

The volume of the base Mg(OH)2 used is 5.28 L

You might be interested in

The lower the ph of a solution, the ______.

gtnhenbr [62]

B

a and c are unrelated
d: ph will increase

3 0

3 years ago

Read 2 more answers

15. Which sample of argon gas has the same number of atoms as a 100.-milliliter sample of helium gas at 1.0 atm and 300. K? A) 5

OLga [1]

The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K

The correct option is D.

<h3>What is the number of moles of gases in the given samples?</h3>

The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.

The ideal gas equation is: PV/RT = n

where;

P is pressure
V is volume
n is number of moles of gas
T is temperature of gas
R is molar gas constant = 0.082 atm.L/mol/K

Moles of gas in the given helium gas sample:

P = 1.0 atm, V = 100 mL or 0.1 L, T = 300 K

n = 1 * 0.1 / 0.082 * 300

n = 0.00406 moles

For the argon gas sample:

A. n = 1 * 0.05 / 0.082 * 300

n = 0.00203 moles

B. n = 0.5 * 0.05 / 0.082 * 300

n = 0.00102 moles

C. n = 0.5 * 0.1 / 0.082 * 300

n = 0.00203 moles

D. n = 1 * 0.1 / 0.082 * 300

n = 0.00406 moles

Learn more about ideal gas equation at: brainly.com/question/24236411

#SPJ1

8 0

1 year ago

If a 250mL beaker weighs 13.473g and the same beaker plus 2.2 mL of water weighs 15.346g. How much does the 2.2 mL of water, alo

Sindrei [870]

Given :

A 250 ml beaker weighs 13.473 g .

The same beaker plus 2.2 ml of water weighs 15.346 g.

To Find :

How much does the 2.2 ml of water, alone, weigh .

Solution :

Now, mass of water is given by :

$\text{Mass of water alone = Total mass - Mass of beaker alone}\\\\Mass =15.346-13.473\ g\\\\Mass =1.873\ g$

Therefore , mass of 2.2 ml of water alone is 1.873 g .

Hence , this is the required solution .

4 0

3 years ago

The study of the composition of matter is?

NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0

3 years ago

The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L/mol/s. What is the instantaneou

pochemuha

Answer: Rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For a reaction : $A\rightarrow products$

$Rate=k[A]^x$

k= rate constant

x = order of the reaction = 2

$Rate=4.71\times 10^{-8}L/mol/s\times {5.55\times 10^{-4}M}^2$

$Rate=1.45\times 10^{-14}$

Thus rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

6 0

3 years ago