1.3 grams of h3po4 is reacted with 4.5 grams of bacl2. if 1.2 grams of ba3(po4)2 is formed, what is the perceng yield of this reaction

Question

2H3PO4 + 3BaCl2 -> Ba3(PO4)2 + 6HCl

Answer 1

2H3PO4 + 3BaCl2 -> Ba3(PO4)2 + 6HCl ... balanced equation

moles H3PO4 present = 1.3 g x 1 mole/97.99 g = 0.013 moles H3PO4

moles BaCl2 present = 4.5 g x 1 mole/208 g = 0.0216 moles BaCl2

Find limiting reactant:

H3PO4: 0.013 moles x 1 mol Ba3(PO4)2/2 mol H3PO4 = 0.0065 moles Ba3(PO4)2 produced

BaCl2: 0.0216 moles x 1 mol Ba3(PO4)2/3 mol BaCl2 = 0.0072 moles Ba3(PO4)2 produced

H3PO4 is LIMITING

From above calculations, the theoretical yield Ba3(PO4)2 = 0.0065 moles

Theoretical mass produced = 0.0065 moles x 602 g/mole = 3.9 g

% yield = actual/theoretical (x100) = 1.2/3.9 (x100) = 30.8% = 31% yield (to 2 significant figures)

Answer 2

Step 1: Make sure the chemical reaction equation is balanced and complete (looks good as it is)

Step 2: Calculate the molar mass of the H₃PO₄

Step 3: Use the molar mass from step 2 to find the number of moles of 1.3 g of H₃PO₄

Step 4: Use the molar ratios in the chemical reaction equation to find the theoretical number of moles of Ba₃(PO₄)₂ formed.

Step 5: Calculate the molar mass of Ba₃(PO₄)₂

Step 6: Use the molar mass from Step 5 to calculate the number of moles in the 1.2 grams of Ba₃(PO₄)₂ that was actually created (will be less than the theoretical)

Step 7: Compare the number of moles actually created (Step 6) to the number of moles that should have been created (Step 4) (Note that dividing them will give you the %)