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Stoichiometry - Mole Ratios & Yield
Chapter summary, hard words and model exam answers for Hindi.
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Chemistry · CBSE 10 · ICSE 10 · GCSE (AQA, Edexcel, OCR)
Summary
In N2 + 3H2 -> 2NH3 the coefficients 1, 3, 2 are fixed. 1 mol N2 always needs 3 mol H2 and gives 2 mol NH3; scale the whole recipe together.
Convert the given mass to moles with n = m/M, apply the mole ratio to find moles of the wanted substance, then convert back to mass with m = n x M.
Divide the moles of each reactant by its coefficient; the smallest value is the limiting reagent. The other reactant is in excess and is left over.
Theoretical yield is the maximum from stoichiometry. Actual yield is lower because of incomplete reactions, side reactions and losses. Percentage yield = (actual/theoretical) x 100%.
Hard words & meanings
| stoichiometry | using the mole ratios in a balanced equation to calculate amounts of reactants and products |
| mole | the chemist's counting unit; one mole is 6.02 x 10^23 particles |
| molar mass | the mass of one mole of a substance in grams, equal to its relative formula mass |
| mole ratio | the ratio of the coefficients in a balanced equation |
| limiting reagent | the reactant that is used up first and limits how much product forms |
| excess | a reactant present in more than enough amount, so some is left over |
| percentage yield | actual yield divided by theoretical yield, times 100 |
| atom economy | the proportion of reactant mass that ends up in the wanted product |
Model exam answers, grammar & audio
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