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AnnyKZ [126]
3 years ago
13

Copper (II) oxide can be heated until it decomposes into elemental copper and oxygen gas. Is 0.695 grams of Copper (II) oxide is

produced from a Copper(II) compound, how many grams of elemental Cu could be produced in the decomposition of CuO?
Chemistry
2 answers:
Oksanka [162]3 years ago
8 0

Answer:

0.555 grams of Cu will be produced

Explanation:

Step 1: Data given

Mass of copper (II) oxide = 0.695 grams

Molar mass Copper (II) oxide = 79.545 g/mol

Step 2: The balanced equation

2CuO → 2Cu + O2

Step 3: Calculate moles CuO

Moles CuO = mass CuO / molar mass CuO

Moles CuO = 0.695 grams / 79.545 g/mol

Moles CuO = 0.00874 moles

Step 4: Calculate moles Cu

For 2 moles CuO we'll have 2 moles Cu and 1 mol O2

For 0.00874 moles CuO we'll have 0.00874 moles

Step 5: Calculate mass Cu

Mass Cu = 0.00874 moles * 63.546 g/mol

Mass Cu = 0.555 grams

0.555 grams of Cu will be produced

insens350 [35]3 years ago
4 0

Answer:

0.55 g of Cu can produced by this decomposition

Explanation:

Let's think the reaction of decomposition:

2CuO → 2Cu + O₂

Ratio is 2:2. 2 moles of CuO can decompose into 2 moles of Cu and 1 mol of oxygen.

Let's convert the mass of oxyde to moles → 0.695 g. 1 mol/ 79.55 g =

8.74×10⁻³ moles

8.74×10⁻³ mol of CuO decompose to 8.74×10⁻³ moles of Cu

Let's find out the mass → 8.74×10⁻³ mol . 63.55 g/1mol = 0.55 g

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Explanation:

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:      26.98    159.69    101.96

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Mass/g:  121          601

===============

Step 2. Calculate the <em>moles of each reactant </em>

Moles of Al         = 121 × 1/26.98

Moles of Al         = 4.485 mol Al

Moles of Fe₂O₃  = 601× 1/159.69

Moles of Fe₂O₃  = 3.764 mol Fe₂O₃

===============

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of Al₂O₃ we can obtain from each reactant.  

<em>From Al </em>

The molar ratio is 1 mol Al₂O₃:2 mol Al

Moles of Al₂O₃ = 4.485 × 1/2

Moles of Al₂O₃ = 2.242 mol Al₂O₃

<em>From Fe₂O₃</em>:

The molar ratio is 1 mol Al₂O₃:1 mol Fe₂O₃

Moles of Al₂O₃ = 3.764 × 1/1

Moles of Al₂O₃ = 3.764 mol Al₂O₃

The <em>limiting reactant</em> is Al because it gives the smaller amount of Al₂O₃.

The <em>excess reactant</em> is Fe₂O₃.

===============

Step 4. Calculate the <em>mass of Al₂O₃ formed </em>

Mass of Al₂O₃ = 2.242 × 101.96

Mass of Al₂O₃ = 229 g Al₂O₃

===============

Step 5. Calculate the <em>moles of Fe₂O₃ reacted </em>

The molar ratio is 1 mol Fe₂O₃:2 mol Al:

Moles of Fe₂O₃ = 4.485 × ½

Moles of Fe₂O₃ = 2.242 mol Fe₂O₃

===============

Step 6. Calculate the <em>moles of Fe₂O₃ remaining </em>

Moles remaining = original moles – moles used

Moles remaining = 3.764 – 2.242

Moles remaining = 1.521 mol Fe₂O₃

==============

Step 7. Calculate the <em>mass of Fe₂O₃ remaining </em>

Mass of Fe₂O₃ = 1.521 × 159.69/1

Mass of Fe₂O₃ = 243 g Fe₂O₃

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Explanation:

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