× – g ∴ mass of mole of atom = . × – g × . × = g ∴ number of moles of element in . kg mole g kg × .
mol g g = mol. . Compound Molecular formula Empirical formula Fructose C H O CH O Caffeine C H N O C H N O 11th Std Chemistry 11th Std Chemistry - - - - . Given 2Al + Fe O → Al O + 2Fe Reactants Products Al Fe O Al O Fe Amount of reactant allowed to react g .
kg - - Number of moles allowed to react mol mol - - Stoichiometric Co-efficient Number of moles consumed during reaction mol mol - - Number of moles of reactant unreacted and number of moles of product formed - mol mol mol Molar mass of Al O formed = mol × g mol – Al O g ( ) ( ) Excess reagent = Fe O Amount of excess reagent left at the end of the reaction = mol × g mol – = g Fe O g ( ) ( ) . Balanced equation for the combustion of ethane C H + O → 2CO + 3H O ⇒ 2C H + 7O → 4CO + H O To produce moles of CO , moles of ethane is required ∴ To produce mole ( g) of CO required 11th Std Chemistry