Chemistry · Some Basic Concepts Of Chemistry · NEET
The LEFTOVER is always the EXCESS reagent. The limiting reagent is fully finished (0 left). The excess reagent is the one you had too much of, so some of it stays behind. So you only calculate leftover mass for the excess reagent, never for the limiting one.
Step 1: Convert both amounts to moles and find the limiting reagent (the one that runs out first). Step 2: Using the balanced equation's mole ratio, find how many moles of the EXCESS reagent got used up by the limiting reagent. Step 3: Leftover moles = starting moles of excess − used moles. Then multiply leftover moles by its molar mass to get grams.
Do the subtraction in MOLES, not grams. First find moles used, subtract from moles you started with, THEN convert the leftover moles to grams at the end. Subtracting grams directly only works by luck when the mole ratio is 1:1 and molar masses match, so it is safer to always subtract in moles.
The limiting reagent controls everything. Take the moles of the limiting reagent, then multiply by the mole ratio from the balanced equation to get moles of excess reagent consumed. Example: for A + B, if the ratio is 1:1 and 0.01875 mol of the limiting reagent reacts, then 0.01875 mol of the excess reagent is consumed.
Moles of NaOH = 1/40 = 0.025 mol. Moles of HCl = 0.025 L × 0.75 M = 0.01875 mol. HCl has fewer moles and the ratio is 1:1, so HCl runs out first, making HCl the limiting reagent and NaOH the excess reagent. HCl uses 0.01875 mol NaOH = 0.75 g, so NaOH left = 1 − 0.75 = 0.25 g = 250 mg.
Always use the coefficients from the balanced equation. For example in N2 + 3H2 → 2NH3, 1 mol N2 reacts with 3 mol H2. If N2 is limiting and 2 mol N2 react, then H2 used = 2 × 3 = 6 mol. Leftover H2 = starting H2 − 6. Never assume 1:1; check the balanced equation.
NEET directly asks this. In 2024 they gave exactly this: '1 g NaOH treated with 25 mL of 0.75 M HCl, mass of NaOH left unreacted.' If you cannot find the limiting reagent and subtract in moles, you cannot answer these stoichiometry questions, which appear almost every year in Some Basic Concepts of Chemistry.
1 gram of NaOH is treated with 25 mL of 0.75 M HCl solution. The mass of NaOH left unreacted is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Mass left = (starting moles of excess reagent − moles consumed) × molar mass. Moles consumed = moles of limiting reagent × (mole ratio from balanced equation). Always subtract in moles, then convert to grams.
No. By definition the limiting reagent is completely used up, so 0 grams of it remain. Only the excess reagent has leftover mass.
Convert every reactant to moles, then divide each by its coefficient in the balanced equation. The reactant with the smallest value is the limiting reagent; the rest are in excess.
Yes. Moles = Molarity × Volume in litres = 0.75 × 0.025 = 0.01875 mol. Remember to change mL to litres by dividing by 1000.
To trick you into unit errors. 0.25 g = 250 mg (multiply grams by 1000). Always match your answer to the units in the options.
