0.443g of silver metal was obtained wen 0.784g of silver nitrate decomposes. What is the percentage yield of the reaction
To calculate the percentage yield of a reaction, you need to compare the actual amount of product obtained with the theoretical amount of product that should have been obtained, and then express it as a percentage.
In this case, the theoretical amount of silver metal can be calculated based on the balanced chemical equation for the decomposition reaction of silver nitrate:
2 AgNO3 -> 2 Ag + O2 + 2 NO2
According to the balanced equation, 2 moles of silver nitrate (AgNO3) decompose to form 2 moles of silver (Ag). The molar mass of AgNO3 is 169.87 g/mol, which means that 0.784 g of AgNO3 is equal to:
0.784 g / 169.87 g/mol = 0.00461 mol of AgNO3
Since the molar ratio between AgNO3 and Ag is 2:2, we can conclude that 0.00461 mol of AgNO3 should theoretically produce:
0.00461 mol of Ag
Now we can calculate the theoretical mass of Ag using its molar mass, which is 107.87 g/mol:
0.00461 mol x 107.87 g/mol = 0.497 g of Ag (theoretical yield)
The percentage yield can be calculated using the formula:
Percentage yield = actual yield / theoretical yield x 100
In this problem, the actual yield of Ag was given as 0.443 g. Plugging the values into the formula:
Percentage yield = 0.443 g / 0.497 g x 100 = 89.12%
Therefore, the percentage yield of the reaction is approximately 89.12%.