Question

Practice #2 - Limiting Reactants in Moles 1. What is the greatest amount of magnesium oxide in moles that can be made with 6.3 mol Mg and 6.9 mol O,? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction? 2 Mg+ 0, à 2 MgO

Answer 1

First to find the limiting reagents we take the ratio i.e given moles / stoichiometric moles.

So we see ratio is less for Mg , so it is a limiting reagents.

Now using stoichiometry comparison we see 6.3 mol of Mg produce 6.3 mole of MgO.

Then excess O2 remain after the reaction is 6.9-6.3 = 0.6 mole

Aus- hen ni 2 mg + ] — 2mgo Limiting agent, calculate the ratio of Given moles and Stochiometrie moles. ... ratsof mg= 6.3 = 3:15 ratio of 0.99 = 3:45 so ratio of my is less, Mg is the limiting agent 2 mol of mg - 2 mole of myo I mole of mg - I mole of migo 6.3 mole of mg 6.3 mole of mgo. 0 0 +1 na age So lim rati is less mo > so maximum moles of Mgo is 6.3 mole. Excess of a lefta after the z reachin. (6.9-6.3) moka o 6 mole. Scanned with CamScanner