The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2 A+4 B \rightarrow 3 C+4 D$, when 5 moles of $A$ react with 6 moles of $B$, then
(a) which is the limiting reagent?
(b) calculate the amount of $C$ formed?
$$2 A+4 B \longrightarrow 3 C+4 D$$
According to the given reaction, 2 moles of $A$ react with 4 moles of $B$.
Hence, 5 moles of $A$ will react with 10 moles of $B\left(\frac{5 \times 4}{2}=10\right.$ moles $)$
(a) It indicates that reactant $B$ is limiting reagent as it will consume first in the reaction because we have only 6 moles of $B$.
(b) Limiting reagent decide the amount of product produced.
According to the reaction,
4 moles of $B$ produces 3 moles of $C$
$\therefore 6$ moles of $B$ will produce $\frac{3 \times 6}{4}=4.5$ moles of $C$.
Match the following.
| A. | 88 g of CO$$_2$$ | 1. | 0.2 mol |
|---|---|---|---|
| B. | $$6.022\times10^{23}$$ molecules of H$$_2$$O | 2. | 2 mol |
| C. | 5.6 L of O$$_2$$ at STP | 3. | 1 mol |
| D. | 96 g of O$$_2$$ | 4. | $$6.022\times10^{23}$$ molecules |
| E. | 1 mole of any gas | 5. | 3 mol |
A. $\rightarrow(2)$
B. $\rightarrow(3)$
C. $\rightarrow$ (1)
D. $\rightarrow$ (5)
E. $\rightarrow(4)$
$$\text { A. Number of moles of } \mathrm{CO}_2 \text { molecule }=\frac{\text { Weight in gram of } \mathrm{CO}_2}{\text { Molecular weight of } \mathrm{CO}_2}=\frac{88}{44}=2 \mathrm{~mol}$$
$$\begin{aligned} \text { B. } \quad 1 \text { mole of a substance } & =N_A \text { molecules }=6.022 \times 10^{23} \text { molecules } \\ & =\text { Avogadro number } \\ & =6.022 \times 10^{23} \text { molecules of } \mathrm{H}_2 \mathrm{O}=1 \mathrm{~mol} \end{aligned}$$
$$\begin{aligned} \text { C. } 22.4 \mathrm{~L} \text { of } \mathrm{O}_2 \text { at } \mathrm{STP} & =1 \mathrm{~mol} \\ 5.6 \mathrm{~L} \text { of } \mathrm{O}_2 \text { at } \mathrm{STP} & =\frac{5.6}{22.4} \mathrm{~mol}=0.25 \mathrm{~mol} \end{aligned}$$
D. Number of moles of 96 g of $\mathrm{O}_2=\frac{96}{32} \mathrm{~mol}=3 \mathrm{~mol}$
E. 1 mole of any gas $=$ Avogadro number $=6.022 \times 10^{23}$ molecules
Match the following physical quantities with units.
| Physical quantity | Unit | ||
|---|---|---|---|
| A. | Molarity | 1. | g ml$$^{-1}$$ |
| B. | Mole fraction | 2. | mol |
| C. | Mole | 3. | Pascal |
| D. | Molality | 4. | Unitless |
| E. | Pressure | 5. | mol L$$^{-1}$$ |
| F. | Luminous intensity | 6. | Candela |
| G. | Density | 7. | mol kg$$^{-1}$$ |
| H. | Mass | 8. | Nm$$^{-1}$$ |
| 9. | kg |
A. $\rightarrow$ (5)
B. $\rightarrow$ (4)
C. $\rightarrow$ (2)
D. $\rightarrow$ (7)
E. $\rightarrow(3)$
F. $\rightarrow(6)$
G. $\rightarrow$ (1)
H. $\rightarrow$ (9)
$$\begin{gathered} \text { A. Molarity }=\text { concentration in } \mathrm{mol} \mathrm{L}^{-1} \\ \text { Molarity }=\frac{\text { Number of } \mathrm{moles}}{\text { Volume in litres }} \end{gathered}$$
B. Mole fraction = Unitless
C. Mole $=\frac{\text { Mass }(\mathrm{g})}{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}=\mathrm{mol}$
D. Molality $=$ concentration in mol per kg solvent
Molality $=\frac{\text { Number of moles }}{\text { Mass of solvent }(\mathrm{kg})}$
E. The SI unit for pressure is the pascal $(\mathrm{Pa})$, equal to one newton per square metre $\left(\mathrm{N} / \mathrm{m}^2\right.$ or $\left.\mathrm{kg} . \mathrm{m}^{-1} \mathrm{~s}^{-2}\right)$. This special name for the unit was added in 1971; before that, pressure in SI was expressed simply as $\mathrm{N} / \mathrm{m}^2$.
F. Unit of luminous intensity = candela.
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540 \times 10^{\text {12 }}$ hertz and that has a radiant intensity in that direction of $1 / 683$ watt per steradian.
G. Density $=\frac{\text { mass }}{\text { volume }}=\mathrm{g} \mathrm{mL}^{-1}$
H. Unit of mass $=$ kilogram
The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram
Assertion (A) The empirical mass of ethene is half of its molecular mass.
Reason (R) The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
Assertion (A) One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason (R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.