NaBR+ AgNO3- AgBr+ NaNO3
AgNO3=reagent A
NaBr=reagent B
When 4.36 grams of AgNO3 react with 5.05 grams of NaBr, which is the limiting reagent?
How many grams of the excess reagent remain?
How many grams of AgBr are produced in the reaction?
To determine the limiting reagent in a chemical reaction, you need to compare the number of moles of each reactant and see which one is less. The reactant that produces fewer moles of the desired product is the limiting reagent.
First, let's calculate the number of moles of AgNO3 and NaBr using their respective molar masses. The molar masses are:
AgNO3: Ag (107.87 g/mol) + N (14.01 g/mol) + 3O (16.00 g/mol) = 169.87 g/mol
NaBr: Na (22.99 g/mol) + Br (79.90 g/mol) = 102.89 g/mol
To find the number of moles, we will use the formula: moles = mass / molar mass.
For AgNO3:
moles AgNO3 = 4.36 g / 169.87 g/mol = 0.0256 mol
For NaBr:
moles NaBr = 5.05 g / 102.89 g/mol = 0.049 mol
Now, let's determine the limiting reagent. We compare the moles of AgNO3 and NaBr and see which one is smaller.
Moles of AgNO3: 0.0256 mol
Moles of NaBr: 0.049 mol
From the calculation, it is clear that AgNO3 has fewer moles than NaBr. Therefore, AgNO3 is the limiting reagent.
To calculate the grams of excess reagent remaining, we need to determine how much of the excess reagent reacted. Since AgNO3 is the limiting reagent, it is completely consumed, and NaBr is in excess.
To find the grams of excess reagent remaining, we need to calculate the grams of NaBr that reacted and subtract it from the initial mass of NaBr.
Grams of NaBr reacted = moles of NaBr (from above calculations) * molar mass of NaBr
Grams of NaBr reacted = 0.049 mol * 102.89 g/mol = 5.04 g
Grams of excess reagent remaining = initial mass of NaBr - grams of NaBr reacted
Grams of excess reagent remaining = 5.05 g - 5.04 g = 0.01 g
Finally, to determine the grams of AgBr produced in the reaction, we need to know the stoichiometric ratio between AgNO3 and AgBr. From the balanced equation, 1 mole of AgNO3 produces 1 mole of AgBr.
Moles of AgBr produced = moles of AgNO3, which is 0.0256 mol (from above calculations)
Grams of AgBr produced = moles of AgBr * molar mass of AgBr
Grams of AgBr produced = 0.0256 mol * (107.87 g/mol + 79.90 g/mol) = 4.84 g
Therefore, the limiting reagent is AgNO3, and 0.01 grams of NaBr remain as the excess reagent. Additionally, 4.84 grams of AgBr are produced in the reaction.