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Question: Given that of \(\vartriangle {H_{comb}}\) cyclopropane is \( - 4000kJmo{l^{ - 1}}\) . The amount of ...

Given that of △Hcomb\vartriangle {H_{comb}} cyclopropane is −4000kJmol−1 - 4000kJmo{l^{ - 1}} . The amount of cyclopropane that needs to be burnt in oxygen for producing 2×105kJ2 \times {10^5}kJ of heat is
A. 20kg20kg
B. 2.1kg2.1kg
C. 21g21g
D. 210mg210mg

Explanation

Solution

The molecular formula for cyclopropane is C3H6{C_3}{H_6} and its molar mass is 42.08g/mol42.08g/mol
Since △Hcomb=−4000kJmol−1Thus,42.08g\vartriangle {H_{comb}} = - 4000kJmo{l^{ - 1}} Thus, 42.08g of cyclopropane has the heat of −4000kJ - 4000kJ.

Complete step by step answer:
Combustion is the chemical process in which a substance reacts rapidly with oxygen and librates the heat. The reactant involved is called fuel and the source of oxygen is called oxidiser. During combustion, the fuel and oxidant react with each other forming different compounds. When cyclopropane undergoes combustion, it breaks down to form water and carbon dioxide with the liberation of energy and the combustion of one mole of cyclopropane can be written as, C3H6+92O2→3CO2+3H2O+4000kJ{C_3}{H_6} + \dfrac{9}{2}{O_2} \to 3C{O_2} + 3{H_2}O + 4000kJ
Since △Hcomb=−4000kJmol−1\vartriangle {H_{comb}} = - 4000kJmo{l^{ - 1}} and thus, the reaction is exothermic.
Thus, one mole of cyclopropane liberates the heat of 4000kJ4000kJ
It is known that one mole of cyclopropane contains 42.08g42.08g Therefore, 42.08g42.08g of cyclopropane liberates the heat of 4000kJ4000kJ
Let “x” be the amount of cyclopropane that needs to be burnt in oxygen for producing 2×105kJ 2 \times {10^5}kJ of heat.
It is known that the molecular formula for cyclopropane is C3H6{C_3}{H_6} and it has three carbons and six hydrogens. Therefore, its molar mass is 42.08g/mol42.08g/mol
The amount of cyclopropane that needs to be burnt in oxygen for producing 2×105kJ2 \times {10^5}kJ of heat can be written as follows,
x=42.08g/mol×2×105kJ4000kJmol−1x = \dfrac{{42.08g/mol \times 2 \times {{10}^5}kJ}}{{4000kJmo{l^{ - 1}}}}
x=0.021×105gx = 0.021 \times {10^5}g
x=0.021×105×10−3kgx = 0.021 \times {10^5} \times {10^{ - 3}}kg
x=0.021×102kgx = 0.021 \times {10^2}kg
x=2.1kgx = 2.1kg

Hence, the correct option is B.

Note: The heat of combustion will be negative since the heat is liberated and the reaction follows an exothermic process.
When we calculate the amount of cyclopropane that needs to be burnt in oxygen for producing 2×105kJ2 \times {10^5}kJ of heat, a negative sign is not needed since mass can’t be negative.