Find the limiting reactant for each initial amount of reactants in the following reaction: 2Na(s)+ Br2(g)→2NaBr(s) A: 5 molNa and 5 molBr2?
2Na(s)+ Br2(g)→2NaBr(s)
If you have 5 mols Br2 that will require 10 mols Na. If you have only 5 mols Na then that is the limiting reagent
Well, let's see. In a balanced chemical equation, the stoichiometric ratios tell us the ideal amounts of each reactant needed for a complete reaction. In this case, we have 2 moles of Na reacting with 1 mole of Br2 to produce 2 moles of NaBr.
To find the limiting reactant, we need to determine which reactant will be completely consumed first and therefore limit the amount of product formed.
Let's do some calculations. If we have 5 moles of Na and 5 moles of Br2, we can use the stoichiometry to determine the amount of product formed by each reactant.
For Na: 5 moles Na × (2 moles NaBr / 2 moles Na) = 5 moles NaBr
For Br2: 5 moles Br2 × (2 moles NaBr / 1 mole Br2) = 10 moles NaBr
So, as you can see, the calculated amount of NaBr formed from Na is 5 moles, while the calculated amount of NaBr formed from Br2 is 10 moles. The reactant that produces the lesser amount of product is the limiting reactant.
Therefore, in this case, Na is the limiting reactant because it produces only 5 moles of NaBr, whereas Br2 could produce 10 moles of NaBr.
To find the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.
The balanced equation for the reaction is:
2Na(s) + Br2(g) → 2NaBr(s)
According to the stoichiometry, 2 moles of Na reacts with 1 mole of Br2 to produce 2 moles of NaBr.
We can calculate the moles of Na reacted by dividing the given amount of Na by its molar mass:
Molar mass of Na = 22.99 g/mol
Moles of Na = 5 mol Na
Moles of Na = (5 mol Na) × (1 mol Na / 22.99 g Na) ≈ 0.22 mol Na
Similarly, we can calculate the moles of Br2 reacted by dividing the given amount of Br2 by its molar mass:
Molar mass of Br2 = 159.81 g/mol
Moles of Br2 = 5 mol Br2
Moles of Br2 = (5 mol Br2) × (1 mol Br2 / 159.81 g Br2) ≈ 0.03 mol Br2
Now, we can compare the moles of each reactant to the stoichiometric ratio.
From the balanced equation, 2 moles of Na react with 1 mole of Br2. Therefore, if we have 0.22 moles of Na, we would need (0.22 moles of Na) / (2 moles of Na) = 0.11 moles of Br2 to react completely.
However, we only have 0.03 moles of Br2, which is less than the required amount of 0.11 moles. Therefore, Br2 is the limiting reactant.
In conclusion, Br2 is the limiting reactant when we have 5 mol of Na and 5 mol of Br2.
To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio given by the balanced chemical equation.
The balanced equation for the reaction is:
2Na(s) + Br2(g) → 2NaBr(s)
From the equation, we can see that the mole ratio between Na and Br2 is 2:1.
For 5 mol Na and 5 mol Br2, we can calculate the moles of Na and Br2 using their molar masses (sodium- 22.99 g/mol, bromine- 79.90 g/mol).
Number of moles of Na = 5 mol × (22.99 g/mol) = 114.95 g
Number of moles of Br2 = 5 mol × (79.90 g/mol) = 399.50 g
Now, let's find the mole ratio of Na to Br2:
Mole ratio of Na to Br2 = (114.95 g Na) / (399.50 g Br2) = 0.288
The stoichiometric ratio for Na to Br2 in the balanced equation is 2:1, which means we need twice as many moles of Na as Br2.
Since the mole ratio of Na to Br2 is less than the required stoichiometric ratio, Na is the limiting reactant.
Therefore, Na is the limiting reactant for the given initial amounts of 5 mol Na and 5 mol Br2.