All categories

3 years ago

What is the theoretical yield of calcium phosphate?

Chemistry

1 answer:

BARSIC [14]3 years ago

5 0

I'm going to assume that the other reactant is sodium phosphate(you did not indicate this)
so i'm guessing that the reaction did this
CaCL2 + Na3(PO4) ------------> Ca3(PO4)2 + NaCl
(since this reaction was done in water(aqueous solutions involved) the NaCl would be in the solution and thus the filtrate of the product so it won't influence the mass of the calcium phosphate product)

so then you need to write the balanced equation for this reaction(lets assume you can do this)
3 CaCL2 + 2 Na3(PO4) ------------> Ca3(PO4)2 + 6 NaCl

so in theoretical analysis 3 moles of calcium chloride is involved in the production of 1 mole of calcium phosphate

in terms of molar mass equivalents 3 molesw of calcium chloride weigh = 3 x (40 + 71)g/mol = 333g
1 mole of calcium phosphate = 310 g
so if you only have 0.513 g of calcium chloride then this would calculate to get the mass of calcium phosphate
mass = [0.513 / 333] x 310 g = 0.477grams

your actual yield is what you massed when you performed the lab. hope that helped.

You might be interested in

Can anyone help me with this asap?

tresset_1 [31]

Answer:

Explanation:

1) During the diagnosis of thyroid disease a 10 g sample of I-131 is used. After a period of 32 days, how much sample is still radio active.?

Answer:

0.625 g

Explanation:

HL = Elapsed time/half life

32 days/8 days = 4

At time zero = 10 g

At 1st half life = 10/2 = 5 g

At 2nd half life = 5/2 = 2.5 g

At 3rd half life = 2.5 /2 = 1.25 g

At 4th half life = 1.25 / 2 = 0.625 g

After 32 days still 0.625 g of I-131 remain radioactive.

2) what was the original mass of sample Tc-99 that was used to locate the brain tumor If 0.10 g of a sample remains after 30 days? (half life 6 days)

Answer:

0.32 g.

Explanation:

Half life = time elapsed / HL

Half life = 30 days / 6 days = 5

At 5th half life = 0.10 g

At 4th half life = 0.2 g

At 3rd half life = 0.4 g

At 2nd half life = 0.8 g

At 1st half life = 0.16 g

At time zero = 0.32 g

The original amount was 0.32 g.

3) write the beta decay equation of I-131?

Equation:

¹³¹I₅₃ → ¹³¹Xe₅₄ + ⁰₋₁e

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiation

¹³¹I₅₃ → ¹³¹Xe₅₄ + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

3 0

3 years ago

Aldehydes plus potassium permanganate with explanation

kozerog [31]

$\large \sf \: answer$

You can identify the aldehyde from acidic potassium permanganate. Purple colour of acidic potassium permanganate is changed to colourless because aldehyde is oxidized by acidic potassium permanganate.

5 0

2 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

A student is given a sample of CuSO4(s) that contains a solid impurity that is soluble and colorless. The student wants to deter

DochEvi [55]

Answer:

The impurity which is present in the solution of copper sulphate (CuSO4) is determined by the an instrument known as spectrophotometer.

Explanation:

Spectrophotometer is a device or an instrument which is used to determine the concentration of a chemical by measuring the detection of light intensity that is coming from the solution. If the solution of copper sulphate is checked through spectrophotometer, we can can determined or measure the amount of copper sulphate and the impurity in the solution.

3 0

3 years ago

Out of sodium & sulfar which is a metal? explain it's reaction with the oxygen.

ANEK [815]

Answer:

sodium is metal because it has one electron in its outer shell and gives it to other to be stable.so it is metal.but sulfur gains 1 electron from other so it is non metal.

7 0

3 years ago

Read 2 more answers