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Citrus2011 [14]
3 years ago
14

A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula

for the hydrate?
Chemistry
2 answers:
Tomtit [17]3 years ago
4 0

Na₂CO₃•H₂O

<h3>Further explanation</h3>

<u>Given:</u>

  • A 35.40 g hydrate of sodium carbonate, Na₂CO₃•nH₂O, is heated to a constant mass.
  • Its final weight is 30.2 g.

<u>Question:</u>

What is the formula for the hydrate?

<u>Problem-solving:</u>

We will solve problems related to The Law of Definite Proportion (Proust's Law).

"The ratio of the masses of elements in each compound is always constant. Put differently, a given compound invariably contains a similar proportion of elements by mass."

The heating reaction of the hydrate of sodium carbonate is as follows:

\boxed{ \ Na_2CO_3.nH_2O \xrightarrow{\text{heat}} Na_2CO_3 + nH_2O \ }

<u>Condition:</u>

  • Before heating: 35.40 g of Na₂CO₃•nH₂O (hydrate)
  • After heating: 30.20 g of Na₂CO₃ (anhydrate)
  • The mass of H₂O that has left the compound is \boxed{ \ 35.40 - 30.20 = 5.2 \ g \ }  

\boxed{ \ Na_2CO_3.nH_2O \xrightarrow{\text{heat}} Na_2CO_3 + nH_2O \ }

         35.4 g                 30.2 g       5.2 g

We want to determine the amount of water (with the symbol n) so that the formula for the hydrate can be known.

Recall that \boxed{ \ Moles = \frac{Mass \ (g)}{Mr} \ }

Let us prepare the number of moles of Na₂CO₃ and H₂O respectively.

  • Mr Na₂CO₃ = 106 g/mol ⇒ \boxed{ \ moles = \frac{30.2}{106} \rightarrow \boxed{ \ 0.285 \ } \ }
  • Mr H₂O = 18 g/mol ⇒ \boxed{ \ moles = \frac{5.2}{18} \rightarrow \boxed{ \ 0.289 \ }\ }

Finally, we use the mole ratio to write the formula. Find the water-to-anhydrate mole ratio.

\boxed{ \ Na_2CO_3 \ : \ H_2O = 1 \ : \ n  \ }

\boxed{ \ H_2O \ : \ Na_2CO_3 = n \ : \ 1  \ }

\boxed{ \ n \ : \ 1 = 0.289 \ : \ 0.285 \ }

\boxed{ \ n = \frac{0.289}{0.285} \times 1 \ }

Thus, the value of n after rounding is \boxed{ \ n = 1 \ }

Substitute n = 1 into Na₂CO₃•nH₂O.

Therefore, the formula for the hydrate is \boxed{\boxed{ \ Na_2CO_3.H_2O \ }}

<h3>Learn more</h3>
  1. The law of multiple proportions (Dalton’s law)  brainly.com/question/10590259  
  2. Conservation of mass  brainly.com/question/9473007
  3. The chemical formula of a compound  brainly.com/question/834909

Keywords: 35.40 gram, hydrate of sodium carbonate, Na₂CO₃•nH₂O, heating, a constant mass, its final weight, 30.2 g, what is the formula for the hydrate? the law of definite proportion, Proust's law, anhydrate, the mole ratio

Sever21 [200]3 years ago
3 0

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1.  So the formula of the hydrate is Na₂CO₃•H₂O.

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3 years ago
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
wlad13 [49]
<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

  • Atoms of Magnesium = 7.179 x 10^23 atoms
  • Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

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Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

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Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

4 0
3 years ago
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