If 80.0 g of bromine is added to 40.0 g of sodium,
a) what is the limiting reactant?
b) what is the excess mass of the reactant?
c) what is the mass of product?
I do these limiting reagent (LR) problems the long way. There are shorter ways to do them.
2Na + Br2 --> 2NaBr
mols Na = grams/atomic mass = ?
mols Br2 = grams/molar mass
Using the coefficients in the balanced chemical equation, convert mols Na to mols NaBr.
Do the same for mols Br2 to mols NaBr.
It is likely these two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR. That goves up a.
c you get by a x molar mass NaBr = g NaBr
b. Using the coefficients in the balanced equation, convert mols of the LR to moles of the OR (other reagent). That gives you the mols of the other reagent used. Then excess mols = initial mols - mols used = final mols.
Convert that to grams by g = mols x atomic mass.
To determine the limiting reactant, excess mass of the reactant, and the mass of the product, we need to perform stoichiometric calculations.
a) The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the amount of product formed. To find the limiting reactant, we need to compare the amount of product that can be formed from each reactant.
First, we need to determine the number of moles of each reactant. To do this, divide the given mass of each reactant by their respective molar masses.
The molar mass of bromine (Br2) is 159.8 g/mol, so the number of moles of bromine is:
80.0 g / 159.8 g/mol = 0.501 mol
The molar mass of sodium (Na) is 22.99 g/mol, so the number of moles of sodium is:
40.0 g / 22.99 g/mol = 1.74 mol
Next, we need to determine the stoichiometric ratio between bromine and sodium in the balanced chemical equation. The balanced chemical equation is required to calculate the amount of product formed.
The balanced chemical equation for the reaction between bromine and sodium is:
2 Na + Br2 → 2 NaBr
From the balanced equation, we can see that for every 1 mole of bromine, 2 moles of sodium bromide (NaBr) are formed.
Since the stoichiometric ratio is 1 mole of bromine to 2 moles of sodium bromide, the number of moles of sodium bromide that can be formed from the available amount of bromine is:
0.501 mol bromine * (2 mol NaBr / 1 mol Br2) = 1.00 mol NaBr
Similarly, we can determine the number of moles of sodium bromide that can be formed from the available amount of sodium:
1.74 mol sodium * (2 mol NaBr / 2 mol Na) = 1.74 mol NaBr
Comparing the two amounts, we can see that 1.00 mol NaBr can be formed from the available bromine, while 1.74 mol NaBr can be formed from the available sodium.
Therefore, the limiting reactant is bromine (Br2) since it produces the lesser amount of product (NaBr).
b) The excess mass of the reactant is the amount of the non-limiting reactant that is left over after the reaction has occurred. To calculate the excess mass, we need to find the difference between the initial mass of the reactant and the mass of the reactant that reacted.
Since sodium is the non-limiting reactant, we can calculate the mass of sodium bromide (NaBr) formed from the given amount of sodium by using the stoichiometric ratio:
1.74 mol NaBr * (22.99 g/mol NaBr) = 40.0 g NaBr
Therefore, the mass of sodium bromide formed from the available sodium is 40.0 g.
To find the excess mass of the sodium reactant, subtract the mass of sodium bromide formed from the initial mass of sodium:
40.0 g - 40.0 g = 0 g
The excess mass of the sodium reactant is 0 g, meaning all of the sodium has reacted and there is none left over.
c) The mass of the product, sodium bromide (NaBr), can be calculated using the stoichiometric ratio between the limiting reactant and the product.
From the stoichiometric ratio in the balanced equation, we know that 2 moles of sodium bromide are formed for every 1 mole of bromine.
Using the limiting reactant (bromine) and its calculated number of moles (0.501 mol), we can calculate the mass of sodium bromide formed:
0.501 mol * (22.99 g/mol NaBr) = 11.5 g
Therefore, the mass of sodium bromide formed is 11.5 g.