Question

For the following reaction, if 9.7 g of AgNO3, is reacted with excess BaCl2, what is the theoretical yield of Ba(NO3)2 for this reaction? Round your answer to the nearest gram.

BaCl2 + 2 AgNO3 → 2 AgCl + Ba(NO3)2

Answer 1

The theoretical yield of Ba(NO3)2 when 9.7 g of AgNO3 is reacted with excess BaCl2 is approximately 7 grams, after calculating moles of reactants and converting to mass.

Step-by-step explanation:

To calculate the theoretical yield of Ba(NO3)2 when 9.7 g of AgNO3 is reacted with excess BaCl2, we first need to determine the moles of AgNO3 using its molar mass, which is 169.87 g/mol. Therefore, we have:

moles of AgNO3 = mass / molar mass = 9.7 g / 169.87 g/mol ≈ 0.0571 mol

According to the balanced reaction equation, each mole of BaCl2 will react with 2 moles of AgNO3 to form 1 mole of Ba(NO3)2. Thus, the moles of AgNO3 will determine the moles of Ba(NO3)2 since AgNO3 is in excess. We can calculate the moles of Ba(NO3)2 as follows:

moles of Ba(NO3)2 = moles of AgNO3 / 2 = 0.0571 mol / 2 ≈ 0.0286 mol

Finally, we convert the moles of Ba(NO3)2 to mass using its molar mass (261.34 g/mol):

mass of Ba(NO3)2 = moles × molar mass = 0.0286 mol × 261.34 g/mol ≈ 7.47 g

The rounded theoretical yield of Ba(NO3)2 is therefore 7 grams.