Question

Maximum mass of sucrose ${{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}$ produced by mixing $\text{84 g}$ of carbon, $\text{12 g}$ of hydrogen and $\text{58 L }{{\text{O}}_{\text{2}}}$ at 1 atm and $\text{273 K}$ according to the given reaction,$\text{ C}\left( \text{s} \right)\text{ + }{{\text{H}}_{\text{2}}}\left( \text{g} \right)\text{ + }{{\text{O}}_{\text{2}}}\left( \text{g} \right)\text{ }\to \text{ }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\left( \text{s} \right)$:
A) $138.5$
B) $155.5$
C) $172.5$
D) $199.5$

Answer 1

The limiting reagent is the reactant which consumes when the chemical reaction is complete. Here, the carbon, hydrogen, and oxygen react to produce the sucrose. The amount of oxygen gas determines the amount of sucrose formed after the reaction.

Complete step by step answer:
Stoichiometry establishes the exact quantitative relationship between the number of moles of reactants and products in the chemical reaction. To exhibit a chemical reaction, the reaction must be balanced in other words each atom on the reactant and product side must be the same. The stoichiometry works on the principle of conservation of mass.
We are given that, the carbon, hydrogen, and oxygen combine to produce the sucrose$\text{ }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }$,
$\text{ C + H + O }\to \text{ }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }$

Let's balance this equation.
$\text{ 24C}\left( \text{s} \right)\text{ + 22}{{\text{H}}_{\text{2}}}\left( \text{g} \right)\text{ + 11}{{\text{O}}_{\text{2}}}\left( \text{g} \right)\text{ }\to \text{ 2}{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\left( \text{s} \right)$
Let's find out the limiting reagent. Limiting reagent is a reactant that is consumed when the reaction is complete. The amount of it determines the amount of product formed. The reaction is balanced. The 24 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms produce the 2 molecules of sucrose$\text{ }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }$.

Let’s first convert the amount into the moles.

1. Moles of carbon (C),
   $\text{moles of C = 84 g }\\!\\!\times\\!\\!\text{ }d\frac{\text{1 mol of C}}{\text{12 g }}\text{ = 7 mole}$

2. Moles of hydrogen $\text{ }{{\text{H}}_{\text{2}}}$ ,
   $\text{moles of }{{\text{H}}_{\text{2}}}\text{ = 12 g }\\!\\!\times\\!\\!\text{ }\dfrac{\text{1 mol of H}}{\text{2 g }}\text{ = 6 mole}$

3. Moles of oxygen$\text{ }{{\text{O}}_{\text{2}}}$,
   $\text{moles of }{{\text{O}}_{\text{2}}}\text{ = 56 of }{{\text{O}}_{\text{2}}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{1 mol of H}}{\text{22}\text{.4 }}\text{ = 2}\text{.5 mole}$
   From the number of moles, it is seen that oxygen is the limiting reagent. The amount of oxygen gas is responsible for product formation.

Let us find out the molecular weight of $\text{ }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }$,
$\text{Mol}\text{.wt of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }=\text{ 12(12) + 22(1) + 11(16) = = 342 g mo}{{\text{l}}^{\text{-1}}}\text{ }$

Since from stoichiometry we know that 11 moles of oxygen give the 2 moles of sucrose, then 2.5 moles of oxygen will give x moles of sucrose.
$\begin{aligned} & \begin{matrix} 11\text{ moles of }{{\text{O}}_{\text{2}}} & = & 2\text{ moles of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}} \\\ 2.5\text{ moles of }{{\text{O}}_{\text{2}}} & = & x\text{ moles of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}} \\\ \end{matrix} \\\ & \text{Then,} \\\ & x\text{ moles of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ = }\dfrac{2.5\text{ }\times \text{ 2}}{11}=\dfrac{5}{11}\text{moles} \\\ \end{aligned}$

Now, the maximum mass-produced by the 2.5 moles of oxygen is,
$\begin{aligned} & \text{Moles of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ = }\dfrac{\text{mass}}{\text{Molar mass of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{ }} \\\ & \text{ }\dfrac{5}{11}\text{ mo}{{\text{l}}^{\text{-1}}}\text{ = }\dfrac{\text{mass}}{\text{342 g mo}{{\text{l}}^{\text{-1}}}} \\\ & \Rightarrow \text{ = }\dfrac{5}{11}\text{mo}{{\text{l}}^{\text{-1}}}\text{ }\times \text{ }342\text{ g mo}{{\text{l}}^{\text{-1}}} \\\ & \therefore \text{mass of }{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}=\text{ 155}\text{.45 g} \\\ \end{aligned}$
Therefore, the mass of sucrose ${{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}$ produced when 84 g carbon, 12 g hydrogen, and 56 litre of oxygen is $\text{155}\text{.45 g}$.
So, the correct answer is “Option B”.

Note: There are various methods to determine the limiting reagent. One of the methods is the guess and check method. Pick up the reagent and pretend it is as the limiting reagent. Now calculate the number of moles of the other reagent .for example, here we will pretend that carbon is the limiting reagent then the amount of oxygen would be,
$\text{moles of }{{\text{O}}_{\text{2}}}=\text{ 7 mole of }\times \text{ }\dfrac{11\text{ moles of }{{\text{O}}_{\text{2}}}\text{ }}{24\text{ mole of }}=3.20\text{ mol}$

Based on these calculations, we would require $3.20\text{ mol}$ oxygen however we have only $\text{2}\text{.5 mole}$of oxygen. Thus carbon is not a limiting reagent.