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nlexa [21]
4 years ago
8

Copper reacts with silver nitrate through single replacement.

Chemistry
1 answer:
-Dominant- [34]4 years ago
6 0

Answer:

  • Part a) 0.0104 moles copper(II) nitrate.

  • Part b)

            i) 0.0418 mole Cu

            ii) 0.0209 mol Ag NO₃

Explanation:

<u>1) Balanced chemical reaction (single replacement):</u>

In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)

  • Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag

<u>2) Mole ratio: </u>

  • 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag

<u />

<u>3) Moles of Ag</u>

  • n = mass in grams / atomic mass
  • atomic mass of Ag: 107.868 g/mol
  • n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag

<u>4) Moles of copper(II) nitrate:</u>

  • Set the proportion using the mole ratio:
  • 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
  • Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ =  0.0104 moles Cu(NO₃)₂

That is the answer of part a: 0.0104 moles copper(II) nitrate.

<u>5) Moles of each reactant</u>

i) Cu:

  • Set a proportion using the theoretical mole ratio

        1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag

  • Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu

ii) Ag NO₃

  • Set a proportion using the teoretical mole ratio

   

       2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag

  • Solve for x: x = 0.0209 mol Ag NO₃
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Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


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3 years ago
If the reactants are 10 kg what is the mass of the products
nlexa [21]

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

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Why do ionic and polar compounds dissolve in water?
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Acetic acid, CH3COOH, can be produced by bubbling oxygen gas into acetaldehyde, CH3CHO, in the presence of
slamgirl [31]

Explanation:

The balanced equation for the reaction is given as;

2CH3CHO + O2 → 2CH3COOH  

If 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel, (a)

how many grams of acetic acid will be produced?

First thing's first, we have to find he limiting reactant. This is done by comparing the number of moles of the reactants.

From the equation of the reaction;

2 mol of CH3CHO reacts with 1 mol of O2

From the masses given;

Number of moles = Mass / Molar mass

CH3CHO;

Number of moles = 20 / 44.0526 = 0.454 mol

O2;

Number of moles = 10 / 32 = 0.3125 mol

The limiting reactant is CH3CHO because O2 would be in excess.

Back to the question;

2 mol of CH3CHO produces 2 mol of CH3COOH  

0.454 mol would produce x

Solving for x;

x = 0.454 * 2 / 2 = 0.454 mol

Converting to mass;

Mass = number of moles* Molar mass

Mass = 0.454 mol *  60.052 g/mol = 27.26 grams

(b) how many grams of the excess reactant remain after the reaction is

complete

The excess reactant is O2

Number of moles left = Initial Number of moles - Number of moles that reacted

Number of moles left =  0.3125 mol - (0.454 mol / 2)

Number of moles left = 0.0855 mol

Converting to mass;

Mass = 0.0855 mol * 32 g/mol = 2.736 grams

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3 years ago
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