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Sever21 [200]
2 years ago
10

uppose the reaction Ca3(PO4)2 + 3H2SO4 ï‚® 3CaSO4 + 2H3PO4 is carried out starting with 153 g of Ca3(PO4)2 and 87.6 g of H2SO4.

How much phosphoric acid will be produced?
Chemistry
1 answer:
aleksklad [387]2 years ago
4 0

The answer is 58.4 g of H₃PO₄

Given that mass of Ca₃(PO₄)₂ is 153g

mass of H₂SO₄ is 87.6g

We need to calculate the mass of H₃PO₄

So the balanced chemical reaction is

Ca₃(PO₄)₂  +  3H₂SO₄  ⇒   3CaSO₄  +  2H₃PO₄

Let us calculate the molar mass of the reactants

Ca₃(PO₄)₂ = (3 x 40) + (2 x 31) + (8 x 16)

= 120 + 62 + 128

= 310 g

H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4)

= 2 + 32 + 64

= 98 g

Now let us calculate the limiting reactant

The Theoretical Yield =  Ca₃(PO₄)₂ / H₂SO₄

= 310 / 3(98)

= 1.05

The Experimental yield

Ca₃(PO₄)₂ / H₂SO₄

= 153 / 87.6 = 1.74

Because the observed percentage was more than the predicted proportion, H2SO4 is the limiting reactant.

Let us Calculate the molar mass of H₃PO₄

H₃PO₄ = (1 x 3) + (31 x 1) + (16 x 4)

= 3 + 31 + 64

= 98 g

Now Calculate the mass of H₃PO₄

3(98) g of H₂SO₄ ------------------ 2(98) g of H₃PO₄

87.6 g of H₂SO₄ ------------------  x

x = ( 87.6 x 2 x 98) / (3 x 98)

x = 17169.6 / 294

x = 58.4g of H₃PO₄

Learn more about Theoretical Yield here

brainly.com/question/25996347

#SPJ1

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2 years ago
Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa
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Answer:

1.  the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c (\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} })

Explanation:

To find  -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 117 kPa

⇒Absolute pressure ( p1 )  = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa ,  Gage pressure = 212 kPa

⇒Absolute pressure (p2)  = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. V_{1} = V_{2}

As we know that, P ∝ M

⇒ \frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }

⇒\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }

⇒\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378 ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒\frac{P}{T} ∝ M

⇒\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }

So, The correct relation is c option.

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Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

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Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

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