Question

Silver bromide is prepared by reacting 200.0 g of magnesium bromide with an adequate amount of silver nitrate. using the following balanced reaction below, what would be the mass of agbr that can be formed?

MgBr₂ + 2AgNO₃ -> Mg(NO₃)₃ + 2AgBr

Answer 1

To find the mass of AgBr that can be formed, we need to calculate the number of moles of AgBr produced and then convert it to grams using the molar mass of AgBr. The balanced equation shows that for every 1 mole of MgBr2, 2 moles of AgBr are produced. By dividing the given mass of MgBr2 by its molar mass and then multiplying by the stoichiometric ratio, we can calculate the number of moles of AgBr. Finally, multiplying the number of moles by the molar mass of AgBr gives us the mass of AgBr that can be formed.

Step-by-step explanation:

To determine the mass of AgBr that can be formed, we need to calculate the number of moles of AgBr produced and then convert it to grams using the molar mass of AgBr.

From the balanced equation, we can see that the stoichiometric ratio between magnesium bromide (MgBr2) and AgBr is 1:2. This means that for every 1 mole of MgBr2, 2 moles of AgBr are produced.

To find the number of moles of AgBr produced, we first need to calculate the number of moles of MgBr2. We can do this by dividing the given mass of MgBr2 (200.0 g) by its molar mass (184.11 g/mol).

Molar mass of AgBr = 187.77 g/mol.

Now, we can calculate the number of moles of AgBr by multiplying the number of moles of MgBr2 by the stoichiometric ratio:

Moles of AgBr = (moles of MgBr2) × (moles of AgBr / moles of MgBr2).

Finally, we can calculate the mass of AgBr by multiplying the number of moles of AgBr by its molar mass:

Mass of AgBr = (moles of AgBr) × (molar mass of AgBr).

Substituting the values:

Moles of AgBr = (200.0 g / 184.11 g/mol) × (2 moles AgBr / 1 mole MgBr2) = 2.168 moles.

Mass of AgBr = (2.168 moles) × (187.77 g/mol) = 407.1 g.