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anzhelika [568]
2 years ago
9

How many moles of FeCl3 could be produced from 6.1 moles of Cly? 2 Fe + 3Cl2 —> 2 FeCl2

Chemistry
1 answer:
Black_prince [1.1K]2 years ago
6 0

Answer:

4.1 moles of FeCl₃

Explanation:

The reaction expression is given as shown below:

         2Fe    +       3Cl₂   →    2FeCl₃

Number of moles of Cl₂  = 6.1moles

So;

  We know that from the balanced reaction expression:

                       3 moles of Cl₂ will produce 2 moles of FeCl₃

Therefore    6.1moles of Cl₂ will produce \frac{6.1 x 2}{3}   = 4.1 moles of FeCl₃

The number of moles is 4.1 moles of FeCl₃

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Can you pls tell me the net ironic equation of H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)
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Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

4 0
3 years ago
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