How do you calculate the mole of reactant and product?

Answer 1

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.

Answer 2

Turn the grams of reactants into moles of reactants using the Periodic Table. For an example, two grams of hydrogen would be equal to two moles of hydrogen because the amu of hydrogen is one gram. Next, you look at your reaction. You can use conversion factors to change between reactants and products. For an example, treat it like a recipe. If your recipe says using three cups of flour will make 12 cookies, and you have six cups (assuming you have enough of everything else) you will be able to make 24 cookies because you have enough for two batches. If you take the recipe parable further, you can set everything equal to everything else. The 3 cups of flour would be equal to the 12 cookies. The different parts of the reaction become conversion factors similar to when you switched between grams and moles.

If you are asking about a limiting reactant reaction, there are a few extra steps. You solve everything out for how much of one product will be created, and then divide by the smallest number. That will give you how many moles of everything will be created.

Answer 3

The question as stated makes no sense. The only way for there not to be a limiting reagent would be to have infinite quantities of all the reactants, in which case you would get infinite quantities of product. Unless "there is no limiting reagent" means "the quantity of the limiting reagent is zero", in which case zero product will be produced. Or, possibly, "there are precisely the needed quantities of each reagent", in which case any one of them can be treated as the limiting reagent and you calculate the amount of product produced normally.

Answer 4

it is the word k

Answer 5

For this the exact reaction equation is an essential keystone (to know which reactants and what products are formed)