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vodka [1.7K]
3 years ago
12

How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose

concentration is 0.3 mol / dm 3
Chemistry
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

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6.0 mol NaOH reacts with
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Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • NaOH: 3 moles
  • H₃PO₄: 1 mole
  • H₂O: 3 moles
  • Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

moles of Na_{3}P O_{4} =\frac{6  moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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