45.1 grams of Sodium Chloride is combined with 72.3 grams of barium phosphate. Balance the following equation
NaCl + Ba3(PO4)2 --> Na3PO4 + BaCl2
How many grams of Barium dichloride will be produced?
To calculate the amount of Barium dichloride (BaCl2) produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that gets completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. The stoichiometric ratio can be obtained by comparing the coefficients of each reactant in the balanced equation.
First, let's calculate the number of moles of each reactant using their respective molar masses:
Molar mass of NaCl:
Na = 22.99 g/mol
Cl = 35.45 g/mol
Total molar mass = 22.99 + 35.45 = 58.44 g/mol
Number of moles of NaCl = mass / molar mass
= 45.1 g / 58.44 g/mol
= 0.772 mol
Molar mass of Ba3(PO4)2:
Ba = 137.33 g/mol
P = 30.97 g/mol
O = 16.00 g/mol
Total molar mass = (3 * 137.33) + (2 * (30.97 + 4 * 16.00)) = 601.91 g/mol
Number of moles of Ba3(PO4)2 = mass / molar mass
= 72.3 g / 601.91 g/mol
= 0.120 mol
Next, we compare the stoichiometric ratio of the reactants to determine the limiting reactant. From the balanced equation, we can see that the ratio of NaCl to Ba3(PO4)2 is 1:1. Therefore, both reactants have a 1:1 stoichiometric ratio.
Since the number of moles of NaCl is greater than the number of moles of Ba3(PO4)2, NaCl is in excess, and Ba3(PO4)2 is the limiting reactant.
Now, we can calculate the number of moles of Barium dichloride (BaCl2) produced using the stoichiometric ratio from the balanced equation. The stoichiometric ratio tells us that for every 1 mole of Ba3(PO4)2, 3 moles of BaCl2 are produced.
Number of moles of BaCl2 = stoichiometric ratio * number of moles of Ba3(PO4)2
= 3 * 0.120 mol
= 0.360 mol
Lastly, we can calculate the mass of BaCl2 using its molar mass:
Molar mass of BaCl2:
Ba = 137.33 g/mol
Cl = 35.45 g/mol
Total molar mass = 137.33 + 2 * 35.45 = 208.23 g/mol
Mass of BaCl2 = number of moles * molar mass
= 0.360 mol * 208.23 g/mol
= 74.96 g
Therefore, approximately 74.96 grams of Barium dichloride (BaCl2) will be produced.