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Fittoniya [83]
1 year ago
8

Give the systematic names for the formulas or give the formulas for the names:

Chemistry
1 answer:
lilavasa [31]1 year ago
4 0

Lead (II) acetate trihydrate is the systematic name for the formula Pb(C₂H₃O₂)₂ . 3H₂O.

<h3>What is Molecular Formula ?</h3>

The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.

<h3>What is Oxidation State ?</h3>

Oxidation state is also known as oxidation number. It is defined as the atom is equal to the total number of electrons which have been removed from the element in order to form chemical bond with other atom.

Now find the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O

Assume the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O be x

x + 2 × (-1) + 3 × 0 = 0

x - 2 + 0 = 0

x = 2

Oxidation state of Pb is +2 or (II)

Thus from the above conclusion we can say that The systematic name for the formula Pb(C₂H₃O₂)₂ .3H₂O is Lead (II) acetate trihydrate.

Learn more about the Molecular Formula here: brainly.com/question/15960587

#SPJ4

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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
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The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

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