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Starting off with masses for each you use the mass-mole relationship n=m/M, where n is the number of moles of a substance (mol), m is the starting mass of the substance (g), and M is the MOLAR mass of the substance (g/mol).

BALANCED REACTION

2NaOH(aq) + CaCl2(aq) ----> 2NaCl + Ca(OH)2

Case 1: Sodium hydroxide is the LIMITING reagent (its molar amount is less than twice the amount of calcium chloride), i.e. NaOH = 5.00 g and CaCl2 = 3.00 g

In this case we use the mass of NaOH to find the number of moles.

n=m/M=5.00g/40.0g/mol=0.125mol

From here we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium).

NaCl:NaOH = 2:2 ratio = 1:1, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.125 mol.

Ca(OH)2:NaOH = 1:2, therefore there will be half as many moles since it takes two moles of reactant to create one mole of product (as dictated by the reaction above): 0.0625 mol

Case 2: Calcium chloride is the LIMITING reagent (its molar amount is less than half the amount of sodium hydroxide), i.e. NaOH = 9.00 g and CaCl2 = 3.00 g

n=m/M=3.00g/111g/mol=0.0270 mol

Just as in Case 1, we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium).

NaCl:CaCl2 = 2:1 ratio, therefore there will be twice as many moles since it takes one mole of reactant to create two moles of product (as dictated by the reaction above): 0.0540 mol

Ca(OH)2:CaCl2 = 1:1 ratio, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.0270 mol.

Keep in mind this only works with a BALANCED chemical reaction.

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How is stoichimetry used to calculate amount of product from amount of reactant?

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