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arlik [135]
3 years ago
15

2C2H2 + 5O2 → 4CO2 + 2 H2O How many grams of CO2 are required to react with exactly 3.00 mol of O2? 106 g 132 g 165 g 76.8 g

Chemistry
1 answer:
cricket20 [7]3 years ago
8 0

Hey there!:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

5 moles O2 ------------- 4 moles CO2

3.00 moles O2 ----------  ( moles of CO2 ?? )

moles of CO2 = 3.00 * 4 / 5

moles of CO2 =  12 / 5

moles of CO2 = 2.4 moles

So, molar  mass CO2 = 44.01 g/mol

Therefore:

1 mole CO2 -------------- 44.01 g

2.4 moles CO2 ---------- ( mass of CO2 )

mass of CO2 = 2.4 * 44.01 / 1

mass of CO2 = 106 g

Answer A


Hope that helps!

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What mass of potassium chloride, KCl, is produced when 12.6 g of oxygen, 02, is produced?
blagie [28]

Mass of KCl= 19.57 g

<h3>Further explanation</h3>

Given

12.6 g of Oxygen

Required

mass of KCl

Solution

Reaction

2KClO3 ⇒ 2KCl + 3O2

mol O2 :

= mass : MW

= 12.6 : 32 g/mol

= 0.39375

From the equation, mol KCl :

= 2/3 x mol O2

= 2/3 x 0.39375

=0.2625

Mass KCl :

= mol x MW

= 0.2625 x 74,5513 g/mol

= 19.57 g

4 0
2 years ago
What volume of a 1.0 M HCl is required to completely neutralize 25.0 ml of a 1.0 M KOH?
kirza4 [7]
The question above can be solved by using this equation: 
CAVA =CBVB
Where:
CA =Concentration of acid = 1.0 M
VA = Volume of acid = ?
CB = Concentration of base = 1.0 M
VB = Volume of base = 25 ml
VA = CBVB / CA
VA = [1 * 25] / 1 = 25 / 1 = 25
VA = 25 ml
Therefore, the volume of acid that is required to completely neutralize the base is 25 ml.<span />
8 0
3 years ago
UGRENT! Please help showing all work
agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

8 0
3 years ago
As electrical energy is converted into heat energy, the total amount in the system
Aleks04 [339]
<span>c. remains the same
</span>----------------------------------------------------
As electrical energy is converted into heat energy, the total amount in the system remains the same
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3 0
3 years ago
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I would go with <span> a molecular compound</span>
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