Question

Which of the following option represent correct limitating reagent in reactions (1),(2)&(3)
(1) C(26g)+O2(20g)gives CO2
(2) N2(60g)+3H2(80g) gives 2NH3
(3)P4(100g)+3O2(200g) gives P4O6​

Answer 1

Answer:

the answer is (2) . n2(60g) + 3h2(80g) gives 2nh3

Answer 2

Answer:

1. Oxygen is limiting reagent
2. Nitrogen is limiting reagent
3. Tetra phosphorus is limiting reagent

Explanation:

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

In a chemical reaction limiting reagent is the reactant that is consumed first and prevents any further reaction from occurring. The amount of product formed during the reaction is determined by the limiting reagent. For example, let us consider the reaction of solution and chlorine. 2Na+Cl2→2NaCl

The determination of the limiting reactant is typically just a piece of a larger puzzle. In most limiting reactant stoichiometry problems, the real goal is to determine how much product could be formed from a particular reactant mixture. The limiting reactant or reagent can be determined by two methods.

• Using the mole ration
• Using the product approach

Number of moles,$n=\frac{Given Mass}{MolecularMass}$

The ratio $\frac{mole}{Sliochiometry Coeff}$ is lowest ie. the limiting reagent

(1) $C(26g)+O_{2}(20g)\rightarrow CO_{2}$

Molecular mass of Carbon = 12g

Molecular mass of Oxygen=16g

Therefore, Number of moles of Carbon=$=\frac{26}{12} =2.17$

Number of moles of  Oxygen $=\frac{20}{32}=\frac{5}{8}$

$C_{2} =$ $\frac{n}{S.C.} =\frac{\frac{13}{6}}{1} =\frac{13}{6}$

$O_{2} =\frac{\frac{5}{8}}{1} =\frac{5}{8}$

$\frac{13}{6} > \frac{5}{8}$

Oxygen is limiting reagent

(2) $N_{2} (60g)+3H_{2} (80g)\rightarrow 2NH_{3}$

Molecular mass of Nitrogen = 28g

Molecular mass of Hydrogen=2g

Number of moles of Nitrogen$=\frac{60}{28} =\frac{15}{7}$

Number of moles of Hydrogen$=\frac{80}{2} =40$

$N_{2} =\frac{n}{S.C.} =\frac{\frac{15}{7}}{1} =\frac{15}{7}$

$H_{2} =\frac{n}{S.C.} =\frac{40}{3}$

$\frac{15}{7} < \frac{40}{3}$

Nitrogen is limiting reagent

(3) $P_{4} (100g)+3O_{2} (200g)\rightarrow P_{4} O_{6}$

Molecular mass of tetra phosphorus = 124g

Molecular mass of Oxygen=32g

Number of moles of tetra phosphorus$=\frac{100}{124} =\frac{25}{31}$

Number of moles of Oxygen$=\frac{200}{32} =\frac{25}{4}$

$P_{4} =\frac{n}{S.C.} =\frac{\frac{25}{31}}{1} =\frac{25}{31}$

$O_{2} =\frac{n}{S.C.} =\frac{\frac{25}{4}}{3} =\frac{25}{12}$

$\frac{25}{31} < \frac{25}{12}$

Tetra phosphorus is limiting reagent

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