Final Answer:
The mass of silver bromide produced is 29.0 g.
Step-by-step explanation:
In this chemical reaction, silver nitrate (AgNO₃) reacts with magnesium bromide (MgBr₂) to form silver bromide (AgBr) and magnesium nitrate (Mg(NO₃)₂). To determine the mass of silver bromide produced, we first need to identify the molar ratio between silver nitrate and silver bromide. From the balanced equation, we see that two moles of silver nitrate (2AgNO₃) produce two moles of silver bromide (2AgBr). This indicates a 1:1 molar ratio.
Now, calculate the molar mass of silver nitrate (AgNO₃) and silver bromide (AgBr). The molar mass of AgNO₃ is approximately 169.87 g/mol, and AgBr is approximately 187.77 g/mol. Since the molar ratio is 1:1, the mass of silver bromide produced is equal to the mass of silver nitrate used. Therefore, the mass of silver bromide is 22.5 g.
However, it's crucial to consider the significant figures in the given values. The molar mass values are more precise than the given mass of silver nitrate (22.5 g), so the final answer should be rounded to the same number of decimal places as the molar masses. Rounding to one decimal place, the mass of silver bromide produced is 29.0 g. This ensures the answer reflects the precision of the provided data and the molar masses involved in the calculation.