sodium burns in chlorine to produce sodium chloride according to the reaction: 2Na(s) + Cl2(g) = 2NaCl(s). A student is asked to predict the mass of sodium chloride produced when 71 g of sodium is placed in a container of 200 g of chorine. The container is heated until the reaction occurs. The student gives the answer as 271 grams. Is this the correct answer? why or why not
You're only choice is to work the problem and see. This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. I work these the long way.
2Na(s) + Cl2(g) = 2NaCl(s)
mols Na = grams/molar mass
mols Cl2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Na to mols NaCl.
Do the same to convert mols Cl2 to mols NaCl.
It is likely that the two values will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR.
Now use the smaller value and convert that to grams NaCl.
g NaCl = mols NaCl x molar mass NaCl.
Compare with the stuent's value.
To determine whether the student's answer is correct, we can calculate the expected mass of sodium chloride produced using stoichiometry.
First, we need to find the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed. To determine the limiting reactant, we compare the given amount of each reactant (in grams) to their molar masses.
The molar mass of sodium (Na) is approximately 23 g/mol, and the molar mass of chlorine (Cl2) is approximately 71 g/mol.
For sodium:
Given mass = 71 g
Molar mass = 23 g/mol
By dividing the given mass by the molar mass, we can find the number of moles of sodium present:
Number of moles of Na = 71 g / 23 g/mol = 3.09 mol
For chlorine:
Given mass = 200 g
Molar mass = 71 g/mol (since Cl2 is a diatomic molecule)
Number of moles of Cl2 = 200 g / 71 g/mol = 2.82 mol
According to the balanced equation, the stoichiometric ratio between sodium (Na) and sodium chloride (NaCl) is 2:2. This means that 2 moles of Na react with 2 moles of Cl2 to produce 2 moles of NaCl.
Now, we need to determine the moles of NaCl that can be formed using the limiting reactant. Since there are 2 moles of Cl2 for every 2 moles of Na, the amount of Cl2 in moles will be equal to the amount of Na in moles.
Therefore, the moles of NaCl formed will be equal to the moles of Cl2:
Moles of NaCl = 2.82 mol
Finally, we need to calculate the mass of NaCl formed. The molar mass of sodium chloride (NaCl) is approximately 58.5 g/mol.
Mass of NaCl = Moles of NaCl x Molar mass of NaCl
= 2.82 mol x 58.5 g/mol
= 165.27 g
According to the calculations, the expected mass of sodium chloride produced is 165.27 grams, not 271 grams as stated by the student.
Therefore, the student's answer is incorrect. The correct answer should be 165.27 grams.