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Chemistry 20 Online
OpenStudy (anonymous):

If 45.0g of Li react with 20.0g N2 according to the equation below, find (A) how many grams of Li3N can be produced and (B) the percent yield if 35.0 g of Li3N were actually obtained. 6Li + N2 --> 2Li3N

OpenStudy (cuanchi):

1) calculate the limiting reactant 45.0g Li x 2/ (6 x MM Li )= X moles of Li3N 20.0g N2 x 2/ MM N2 = Y moles of Li3N X<Y => Li is the limiting reactant X<Y => N2 is the limiting reactant 2) to calculate A) multiply the moles of Li3N produced by the limiting reactant by the MM of Li3N 3) to calculate B) divide (35.0 g of Li3N x 100) by the grams of Li3N produced by the limiting reactant calculated in the step 2

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