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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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If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make
vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

              2H₂  + O₂  -----------> 2H₂O

Now look at the reaction for mole ratio

             2H₂  + O₂  -----------> 2H₂O

             2 mole                       2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂  = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

                  2H₂         +       O₂        ----------->    2H₂O

          2 mole (2 g/mol)                                 2 mole (18 g/mol)

                    4 g                                                     36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

                       4 g of H₂ ≅ 36 g of H₂O

                        30 g of H₂ ≅ X of H₂O

Do cross multiplication

                  X of H₂O =  30 g x 36 g / 4 g

                  X of H₂O =  270 g

Now convert grams of H₂O into moles

               No. of moles = mass in grams/molar mass

Put values in above formula

               No. of moles = 270 g / 18 (g/mol)

               No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

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