0.658 g of potassium bromate and 0.562 g of silver nitrate are added to 388 mL water. Solid silver bromate is formed, dried, and weighed. What is the mass, in g, of the precipitated silver bromate? Assume silver bromate is completely insoluble.
I've calculated the moles and have this equation so I know it is a 1 to 1 ratio...but I'm completely stumped now:
KBrO3 + AgNO3 ----> KNO3 + AgBrO3
That means that AgNO3 is the limiting reactant as it has 0.0033 moles compared to 0.0039 for silver nitrate.
But I don't get the last part you said. What same number of moles of silver bromate would I get? Do you mean same number of moles as the limiting reactant? Meaning that there would be 0.0033 moles of silver bromate?
To find the mass of the precipitated silver bromate, we need to determine the limiting reagent first.
1. Convert the masses of potassium bromate (KBrO3) and silver nitrate (AgNO3) to moles.
Molar mass of KBrO3:
K (potassium) = 39.10 g/mol
Br (bromine) = 79.90 g/mol
O (oxygen) = 16.00 g/mol
3O atoms = 3x(16.00 g/mol) = 48.00 g/mol
Total molar mass = 39.10 g/mol + 79.90 g/mol + 48.00 g/mol = 166.00 g/mol
Moles of KBrO3 = mass / molar mass
Moles of KBrO3 = 0.658 g / 166.00 g/mol = 0.00397 mol
Molar mass of AgNO3:
Ag (silver) = 107.87 g/mol
N (nitrogen) = 14.01 g/mol
3O atoms = 3x(16.00 g/mol) = 48.00 g/mol
Total molar mass = 107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol
Moles of AgNO3 = mass / molar mass
Moles of AgNO3 = 0.562 g / 169.88 g/mol = 0.00331 mol
2. Determine the stoichiometric ratio between KBrO3 and AgBrO3 from the balanced chemical equation.
From the given balanced equation:
KBrO3 + AgNO3 → KNO3 + AgBrO3
The ratio between KBrO3 and AgBrO3 is 1:1.
3. Compare the moles of KBrO3 and AgNO3 to determine the limiting reagent.
Since the ratio is 1:1, the moles of KBrO3 and AgNO3 are equal.
4. Convert the moles of AgBrO3 to mass using the molar mass of AgBrO3.
Molar mass of AgBrO3:
Ag (silver) = 107.87 g/mol
Br (bromine) = 79.90 g/mol
3O atoms = 3x(16.00 g/mol) = 48.00 g/mol
Total molar mass = 107.87 g/mol + 79.90 g/mol + 48.00 g/mol = 235.77 g/mol
Mass of AgBrO3 = moles × molar mass
Mass of AgBrO3 = 0.00331 mol × 235.77 g/mol ≈ 0.779 g
Therefore, the mass of the precipitated silver bromate is approximately 0.779 grams.
To find the mass of the precipitated silver bromate, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed.
Step 1: Calculate the moles of each reactant.
Given:
Mass of potassium bromate (KBrO3) = 0.658 g
Mass of silver nitrate (AgNO3) = 0.562 g
To calculate the moles of each reactant, divide the mass by their respective molar masses. The molar masses are:
Molar mass of KBrO3 = 39.10 g/mol + 79.90 g/mol + 16.00 g/mol × 3 = 167.01 g/mol
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + 16.00 g/mol × 3 = 169.87 g/mol
Moles of KBrO3 = 0.658 g / 167.01 g/mol
Moles of AgNO3 = 0.562 g / 169.87 g/mol
Step 2: Determine the limiting reactant.
The balanced equation shows that the reaction occurs in a 1:1 mole ratio. Therefore, the reactant that has the smaller number of moles will be the limiting reactant.
In this case, it appears that silver nitrate (AgNO3) has fewer moles.
Step 3: Calculate the moles of silver bromate (AgBrO3) formed.
Since the limiting reactant is silver nitrate (AgNO3), the moles of silver bromate formed will be equal to the moles of AgNO3 used.
The balanced equation shows that the stoichiometric ratio between AgNO3 and AgBrO3 is 1:1.
Moles of AgBrO3 = Moles of AgNO3 = Moles of KBrO3 (since they are present in a 1:1 ratio)
Step 4: Calculate the mass of AgBrO3 formed.
To calculate the mass of AgBrO3, multiply the moles of AgBrO3 by its molar mass.
Molar mass of AgBrO3 = 107.87 g/mol + 79.90 g/mol + 16.00 g/mol × 3 = 304.77 g/mol
Mass of AgBrO3 = Moles of AgBrO3 × Molar mass of AgBrO3
Now, you can plug in the values into the equation and calculate the mass of the precipitated silver bromate.
You have to calculate the moles of the potassium bromate and silver nitrate from the masses.
Then, take the LESSOR of these as the limiting reactant. There is a 1:1 ratio, so you get the same number of moles of silver bromate. Convert to grams.