Question

Equal weights of X and Y are reacted to form the compound $X_2{Y}_3$. Which of the following statements holds true?
(Molar mass of X = 36 g/mol, molar mass of Y = 24 g/mol)

• A. X is the limiting reagent
• B. Y is the limiting reagent
• C. No reactant is left over and mass of $X_2{Y}_3$ formed is double the mass of ‘X’ taken
• D. None of these

Answer 1

The correct option is C No reactant is left over and mass of $X_2{Y}_3$ formed is double the mass of ‘X’ taken

$2X+3Y\to X_2{Y}_3$
Molar mass of $X_2{Y}_3=144$ g/mole
Since, mass of each reactant is given as equal let us consider 1 g of each reactant is present.
Stoichiometric moles of X = 2 moles
Mass of X reacting = $2\times 36$ = 72 g
Stoichiometric moles of Y = 3 moles
Mass of Y reacting = $3\times 24$ = 72 g
As per stoichiometry,
2 moles of X form 1 mole of $X_2{Y}_3$
72 g of X form 144 g of $X_2{Y}_3$
1 g of X will form = $\frac{144}{72}$ = 2 g of $X_2{Y}_3$
Similarly for Y,
72 g of Y forms 144 g of $X_2{Y}_3$
1 g will form = $\frac{144}{72}$ = 2 g
So, both the reactants will be consumed.
Mass of the product formed is double than that of the reactant X.