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

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Answer 2

To determine the theoretical yield of Ba(NO3)2, we need to calculate the molar masses of AgNO3 and Ba(NO3)2 and use stoichiometry to determine the molar ratio between AgNO3 and Ba(NO3)2.

Molar mass of AgNO3:
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (3 oxygen atoms)
Total molar mass = 107.87 + 14.01 + (16.00 * 3) = 169.87 g/mol

Using the molar mass of AgNO3, we can calculate the number of moles present in 9.7 g:
moles of AgNO3 = mass / molar mass = 9.7 g / 169.87 g/mol

Since the reaction stoichiometry tells us that the molar ratio between AgNO3 and Ba(NO3)2 is 2:1, we can determine the number of moles of Ba(NO3)2 that will be produced.

moles of Ba(NO3)2 = (moles of AgNO3) / 2

Finally, to find the theoretical yield of Ba(NO3)2, we multiply the moles of Ba(NO3)2 by its molar mass.

Theoretical yield of Ba(NO3)2 = (moles of Ba(NO3)2) * molar mass of Ba(NO3)2

Given that Ba(NO3)2 has a molar mass of:
Ba = 137.33 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (6 oxygen atoms)
Total molar mass = 137.33 + 14.01 + (16.00 * 6) = 261.34 g/mol

Now we can calculate the theoretical yield:

moles of AgNO3 = 9.7 g / 169.87 g/mol
moles of Ba(NO3)2 = (moles of AgNO3) / 2
theoretical yield of Ba(NO3)2 = (moles of Ba(NO3)2) * molar mass of Ba(NO3)2

After calculating the above values, round the final answer to the nearest gram to obtain the theoretical yield of Ba(NO3)2.