Stoichiometry-Exercise Question Answers

 

Chemistry

STOICHIOMETRY

EXERCISE SHORT QUESTIONS

Q1.    58.5 amu are termed as formula mass and not the molecular mass of NaCl. Why?

Because the term molecule is used for the substances having covalent bonds whereas NaCl is an ionic substance and molecules of an ionic substance are termed as formula units, not molecules. That is why 59.5 amu is the formula mass of NaCl and not the molecular mass.

Q2     Concept of limiting reactant does not apply to the reversible reactions, why?

A limiting reactant is a reactant that is completely consumed in a chemical reaction. Whereas in a reversible reaction, no reactant is 100% consumed but is regenerated due to a reversible reaction. Therefore the concept of limiting reactant does not apply to a reversible reaction.

Q3.    Differentiate between limiting and non-limiting reactants?

Limiting reactant:
The reactant is present in lesser amounts and consumed first and controls the amount of product formed in a chemical reaction.
 
Non-Limiting Reactant:
The reactant is present in excess amount and some amount of that reactant is left un-reacted on completion of the reaction.

Q4.    What are stoichiometry amounts?

The branch of chemistry deals with the quantitative relationship between reactants and products in a balanced chemical equation.
Calculation based on balanced chemical equations is called stoichiometric amounts. 
For example:            N2 + 3H2 󠀲󠀳󠀴→ 2NH3 
The equation tells that 1 mole of N2(28g) reacts with 3 moles of H2(6g) to produce 2 moles of NH3(34g). Here, 1 mole of N2(28g), 3 moles of H2(6g), and 2 moles of NH3(34g) are stoichiometric amounts.

Q5.    What are representative particles in one mole of a gas at S.T.P? 

According to Avogadro's law, these are 6.02 X 1023 particles present in 22.414 dm3 of a gas.

Q6.   The actual yield is less than the theoretical yield. Give reasons?

Actual yield is often less than theoretical yield due to the following reasons.
  1. Impurities in reactants 
  2. Side reactions
  3. Some reversible reaction
  4. A mechanical loss like wastage during filtration and transfer of products from one container to another container.
  5. Human error

Q7.   What is the relationship between mass and volume of a gas at S.T.P?

1 mole of any gas at S.T.P occupies a volume of 22.414 dm3. (1dm3 = 1 Litre)

Q8.   What is the conversion factor?

Mole ratios of reactant and product as given by balanced chemical equation is called conversion factor. 
For example:  C3H8 + 5O2  󠀲󠀳󠀴→  3CO2 + 4H2O
The equation tells that one mole of propane reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H20. This mole ratio is called the conversion factor. The conversion factor is not affected by the amount of reactant and products.

Q9.   Define theoretical yield and actual yield?

Theoretical yield:
The amount of product calculated from a balanced chemical equation is called theoretical yield.
Actual yield:

The amount of product obtained experimentally as a result of the chemical equation is called actual yield.

Q10.  What is a limiting reactant? How would you determine it?

The limiting reactant is the reactant which is a lesser amount and controls the amount of product formed in a chemical reaction.
Determination of limiting reactant: 
To identify a limiting reactant following 3 steps are performed.
  1. Calculate the number of miles of each given amount of reactant.
  2. Find the number of moles of each reactant taking part in a balanced chemical equation.
  3. Identify the reactant which produces the least amount of product. And that is the limiting reactant.

Q11.  How many molecules of water are there in 12g of ice?

\[1\;  mole\;  of\;  ice (H_{2}O) = 18.016g\]
According to Avagadros's law
\[18.016g\; of\; H_{2}O = 6.02 X 10^{23}\; molecules\]
\[1g\; of\; H_{2}O = \frac{6.02 X 10^{23}}{18.016}\]
\[12g\; of\; H_{2}O = \frac{6.02 X 10^{23}}{18.016}\; X\; 12  =  4\; X\; 10^{23}\; molecules\]

Q12.  How many covalent bonds are present in 9g of H2O?

\[mass\; of\: H_{2}O\: =\: 9g\]
\[Molar\; mass\; of\; H_{2}O\; =\; 18 amu\]
\[No\; of\; moles\; of\; H_{2}O\; =\; \frac{9}{18}\; =\; 0.5\; moles\]
\[1\; molecule\; of\; H_{2}O\; =\; 2\; covalent\; bonds\]
\[And\; 1\; mole\; of\; H_{2}O\; =\; 6.02\; X\; 10^{23}\; X\; 2\; covalent\; bonds\]
\[0.5\; moles\; of\; H_{2}0\; =\; 0.5\; X\; 6.02\; X\; 10^{23}\; X\; 2\; covalent\; bonds\]
\[0.5\; moles\; of\; H_{2}0\; =\; 6.02\; X\; 10^{23}\; covalent\; bonds\]

Q13.  One mole of H2SO4 should completely react  with 2 moles of NaOH. How does Avagadro's number help to explain it?

\[H_{2}SO_{4}\; +\; 2NaOH\; \rightarrow\; Na_{2}SO_{4}\;+\;2H_{2}O\]
During this reaction, total numbers of molecules of reactant react and produce new molecules of products. And according to above reaction 1 mole or 6.02 X 1023 molecules of H2SO4 react with 2 moles or 1.204 X 1024 molecules of NaOH and produce 1 mole of Na2SO4 or 6.02 X 1023 molecules of Na2SO4 and 2 moles or 1.204 X 1024 molecules of H2O.

Q14.  Give reasons that moles of different compounds have different masses but have the same no. of molecules?

Avagadro's determined and counted that 1 mole of a substance has 6.02 X1023 molecules. As different molecules have different atoms having different atomic masses, so their molecular masses may be different but the number of molecules in 1 mole must be the same i.e. 6.0 X 1023.

Q15. 23g of Na sodium is 238g of uranium have equal no of atoms in them?
The atomic mass of sodium is 23 and that of uranium is 238. It means 23g of Na is 1 mole of Na and 238g of U is 1 mole of U. According to Avagadro's law, 1 mole of any element has 6.02 X 1023
atoms. It means that 23g of Na and 238g of U have an equal number of atoms in them i.e. 6.02 X 1023.

Q16. What is percentage composition? Calculate the percentage composition of glucose C6H12O6?

The percentage of an element by mass in the formula mass of a substance is called its percentage composition. We can calculate the percentage composition of each element in glucose by using this formula.
\[Percentage\; of\; an\; element\; =\; \frac{Mass\; of\; element}{Molar\; mass\; of\; compound}\; X\; 100\]
Percentage composition of glucose:
\[Formula\; mass\; of\; Glucose\; C_{6}H_{12}O_{6}\; =\; (6\;X\;C)\; +\; (12\;X\;H)\; +\;(6\;X\;O)\]
\[Formula\; mass\; of\; Glucose\; C_{6}H_{12}O_{6}\; =\; (12\;X\;6)\; +\; (12\;X\;1.008)\; +\; (6\;X\;16)\]
\[Formula\; mass\; of\; Glucose\; C_{6}H_{12}O_{6}\; =\; 180.096\; amu\]
Now;
\[Percentage\; of\; C\; =\; \frac{72}{180.096}\; X\; 100 \;=\; 39.978 \; percent\]
\[Percentage\; of\; H\; =\; \frac{12.096}{180.096}\; X\; 100 \;=\; 6.716 \; percent\]
\[Percentage\; of\; O\; =\; \frac{96}{180.096}\; X\; 100 \;=\; 53.305 \; percent\]

Q17.  Calculate the weight of oxygen gas evolved when 5.0g of KClO3 are completely decomposed thermally?

Given mass of KClO3 = 5g
Formula mass of KClO3 = 39 + 35.5 + 48
                                         = 122.5 amu
Number of moles of KClO3 = (5/122.5) = 0.04 moles
Now;
\[2KClO_{3}\; \xrightarrow[]{\Delta } \;2KCl \; + \; 3O_{2}\]
According to equation
2 moloes of KClO3     =    3 moles of O2
1 mole of KClO3         =    (3/2) moles of O2
0.04 moles of KClO3   =    (3/2) x 0.04 moles of O2
                                     =    0.06 moles of O2
Mass of O2 in gram     =    0.06 x 32
                                     =    1.92 grams

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