36 g of A reacts with 54 g of B to form AB\(_2\), if the molar mass of A and B is respectively 60 and 80, then choose the correct option from the following.
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To determine the limiting reagent, calculate the number of moles of each reactant and compare the ratios based on the balanced equation.
To solve the problem of determining the limiting reagent in the reaction between 36 g of A and 54 g of B to form AB\(_2\), we need to follow these steps:
Calculate the number of moles of each reactant using their given masses and molar masses:
Number of moles of A = \(\frac{36 \text{ g}}{60 \text{ g/mol}}\) = 0.6 moles
Number of moles of B = \(\frac{54 \text{ g}}{80 \text{ g/mol}}\) = 0.675 moles
Determine the stoichiometric ratio of A to B in the compound AB\(_2\):
The balanced chemical equation is A + 2B → AB\(_2\)
Find the limiting reagent:
According to the equation, 1 mole of A requires 2 moles of B. For 0.6 moles of A, the required moles of B = \(2 \times 0.6 = 1.2 \text{ moles}\).
Actual moles of B available = 0.675 moles.
Because 0.675 < 1.2, B is the limiting reagent and will determine the amount of product formed.
Evaluate the amount of product AB\(_2\) formed:
Using B as the limiting reagent, 0.675 moles of B can react completely with half its quantity (as per the stoichiometry) of A to form AB\(_2\).
Number of moles of AB\(_2\) formed = 0.675 / 2 = 0.3375 moles, because 2 moles of B form 1 mole of AB\(_2\).
Molar mass of AB\(_2\) = 60 + 2 \times 80 = 220 g/mol.
Mass of AB\(_2\) formed = 0.3375 moles × 220 g/mol = 74.25 g.
Based on these calculations, the limiting reagent is B. Thus, the correct answer is: Limiting Reagent is B.
