If 40g of Sodium (Na) reacts with 60g of Chlorine (Cl2), what is the mass of the products?
The reaction between Sodium (Na) and Chlorine (Cl2) can be described by the following balanced equation:
2Na + Cl2 → 2NaCl
To determine the mass of the products, we need to check whether any of the reactants is a limiting reactant, meaning it will determine how much product is formed.
First, we need to determine the number of moles of each reactant. To do this, we will use the molar mass of each element:
Molar mass of Na: 22.99 g/mol
Molar mass of Cl2: 2 x 35.45 g/mol = 70.90 g/mol
Number of moles of Na: 40 g / 22.99 g/mol = 1.74 mol
Number of moles of Cl2: 60 g / 70.90 g/mol = 0.847 mol
Now let's check the mole ratio between Na and Cl2 using the balanced equation:
2 moles of Na: 1 mole of Cl2
If we divide the number of moles by the stoichiometric coefficients from the balanced equation, we can find the limiting reactant:
1.74 mol Na / 2 = 0.87
0.847 mol Cl2 / 1 = 0.847
Na has a larger result (0.87), so Cl2 is the limiting reactant.
Now, let's use this information to determine the mass of the products formed:
Since 1 mole of Cl2 reacts with 2 moles of Na, and we have 0.847 moles of Cl2, then we will have 2 x 0.847 = 1.694 moles of NaCl formed.
The molar mass of NaCl is 58.44 g/mol (22.99 + 35.45). Thus, the mass of NaCl formed is:
1.694 mol NaCl x 58.44 g/mol = 98.9 g
Therefore, the mass of the products after the reaction is 98.9 grams.