KPK 11th Class Chemistry Chapter 1 Stochiometry Short Questions Answers

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Class/Subject: 11th Class Chemistry

Chapter Name: Stochiometry

Board: All KPK  Boards

• Malakand Board 11th Class Chemistry Chapter 1 Stochiometry  short questions Answer
• Mardan Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Peshawar Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Swat Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Dera Ismail Khan Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Kohat Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Abbottabad  Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer
• Bannu Board 11th Class Chemistry Chapter 1 Stochiometry short questions Answer

• All KPK Boards 11th Class  Chemistry Annual Examination
• Schools 11th Class Chemistry December Test
• KPK 11th Class Chemistry Test
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KPK 11th Class Chemistry Chapter 1 Stochiometry Short Questions Answers

What is gram atom? Why the concept of gram atom is useful in chemistry?

Gram atom or mole: the atomic mass of an element expressed in grams is called gram atom mass of the element. It is also called gram atom.e.g.
Gram atomic mass of Hydrogen = 1g
Gram atomic mass of Oxygen = 16g
Gram atomic mass of Chlorine = 35g
In chemistry it is used in cal calculations for example. Mole – mass conversion mole – mole conversion mass –mole conversion.

The mass of 5 moles of an element X is 60g. calculate the molar mass of this element name the element.

Solution:
Mass of the element (given = 60g
No of moles of the element X = 5
Molar mass of the element X =?
No of moles of the element X=mass of the element X/molar mass of the element X
Putting the values we get
5 = 60g/molar mass of the element X
Molar mass of the element = X = 60 60/5 = 12
The element X is Carbon

Explain why balanced chemical equations are used in stoichiometric problems?

A balanced chemical equation indicates the number of moles of reaction and products involved in the reaction. We can use the idea of the mole to find reaction or products mass from the equation for the reaction. There are various ways of doing either in moles or mass units. As such – mole, mole-mass and mass –mass. Volumes units in case of gases are also used in these problems. Two basic assumptions are made while doing these calculations.
Reactions are completely converted into products.
No side reactions take place

How will you identify the limiting reagent in a reaction?

Following steps should be taken while finding and calculating the limiting reagent in a chemical reaction.
Balance the chemical equation for the chemical reaction.
Covert the given information into moles.
Calculate the mole ration from the given information.
Compare the calculated ratio to the actual ration.
Use stoichiometry for each individual reaction to find the mass of product produced.
The reaction that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess regent.
To find the amount of remaining excess reactant, subtract the mass of excess regent consumed form the total mass of excess reagent given

Define molar volume of a gas molar volume of a gas. What will be the volume of 2.5 moles of gas H2 and 60 of NH3 at STP?

At S.T.P one mole of any gas occupies 22.4 dm3. This is called molar volume) e.g.
1 mole of O2 = 22.4 dm3 at S.T.P = 32g of  O2 = 6.023 x 1023 molecules of O2
I mole of CO2 = 22.4 dm3 at S.T.P = 44g of  CO2 = 6.023 x 1023 molecules of CO2
Volume of 2.5 moles of H2 gas at STP = ?
1 mole of H2 gas at STP occupies = 22.4 dm3 at STP
2.5 moles of H2 at STP occupy = 22.4 x 2.5 = 56 dm3
Volume of 60g of NH3 at STP =?
No of moles of the NH3 = mass of the NH3/molar mass of NH3
Molar mass of NH3 = 1 x 24 + x 16 =17g/mol
Putting the values we get
No of moles of the NH3 = 60g/ 17g/mol = 3.53
1 mole of NH3 gas at STP occupies = 22.4 dm3
2.5 moles of NH3 at STP occupy = 22.4 x 3.53 = 79.1 dm3

Why the actual yield is usually less than the theoretically yield?

The actual yield is less than the theoretical yield due to the following reasons.
Reactions are reversible. So they do not proceed 100 percent from left to right.
Mechanical loss of the product – taking place during the course separation and purification.
By product produced by the side reaction:
Reaction conditions like temperature, pressure, concentration etc might have disturbed.
It is therefore desirable for a chemist, to check and express the efficiency of a chemical reaction in terms of percentage yield which is in fact a comparison of the actual yield and the theoretical yield.

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