Final answer:
Using stoichiometry, we calculate that the reaction between 0.25 moles of magnesium bromide and 5.6 grams of silver nitrate will yield 93.885 grams of silver bromide and 37.0775 grams of magnesium nitrate.
Step-by-step explanation:
Firstly, we should write down the balanced chemical equation for the reaction between magnesium bromide (MgBr2) and silver nitrate (AgNO3). The likely reaction based on the provided information would be a double replacement reaction, yielding silver bromide (AgBr) and magnesium nitrate (Mg(NO3)2) as products:
MgBr2 + 2AgNO3 → 2AgBr + Mg(NO3)2
Now, calculate the molar mass of silver nitrate, which is 169.87 g/mol. Using this molar mass, determine the number of moles of silver nitrate present:
Number of moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3
= 5.6 g / 169.87 g/mol
= 0.033 moles of AgNO3
Magnesium bromide is the limiting reagent because we have 0.25 moles of it and the reaction requires 2 moles of silver nitrate for each mole of magnesium bromide. Therefore, not all the silver nitrate will react. To find out how much silver bromide and magnesium nitrate will form, we perform stoichiometric calculations.
For silver bromide (AgBr), the stoichiometric mole ratio to magnesium bromide in the balanced equation is 2:1. Hence, we can calculate the moles of silver bromide formed by using the number of moles of the limiting reagent, magnesium bromide:
Moles of AgBr formed = 2 x Moles of MgBr2 (limiting reagent)
= 2 x 0.25 moles
= 0.50 moles
The molar mass of silver bromide is approximately 187.77 g/mol. Thus, the mass of silver bromide formed will be:
Mass of AgBr = Moles of AgBr x Molar mass of AgBr
= 0.50 moles x 187.77 g/mol
= 93.885 grams of silver bromide
To calculate the mass of magnesium nitrate formed, we use the fact that 1 mole of magnesium bromide yields 1 mole of magnesium nitate:
Moles of Mg(NO3)2 formed = Moles of MgBr2
= 0.25 moles
The molar mass of magnesium nitrate is approximately 148.31 g/mol. Thus, the mass of magnesium nitrate formed will be: Mass of Mg(NO3)2 = Moles of Mg(NO3)2 x Molar mass of Mg(NO3)2
= 0.25 moles x 148.31 g/mol
= 37.0775 grams of magnesium nitrate