Final answer:
To balance the given equation (Na(g) + Cl(g) -> NaCl(g)) and calculate the volume of chlorine gas required for the complete reaction of 4.81 g of sodium, follow these steps: 1. Balance the equation by adding a coefficient of 2 in front of NaCl. 2. Convert the mass of sodium to moles. 3. Use the coefficients of the balanced equation to find the mole ratio between sodium and chlorine. 4. Apply the molar volume of gases at STP (22.4 L/mol) to calculate the volume of chlorine gas required.
Step-by-step explanation:
To balance the given equation:
Na (g) + Cl (g) → NaCl (g)
We need to ensure that the number of atoms of each element on both sides of the equation is equal.
For this equation, we have 1 Na on the left side and 1 Na on the right side, so that is balanced.
We also have 1 Cl on the left side, but 1 C on the left side and 1 Cl on the right side, so this is not balanced.
To balance the chlorine atoms, we need to add a coefficient of 2 in front of the NaCl on the right side:
Na (g) + Cl (g) → 2NaCl (g)
Now, the equation is balanced.
To calculate the volume of chlorine gas required for the complete reaction of 4.81 g of sodium, we need to use the balanced equation and convert the mass of sodium to moles. Then, we use the coefficients of the balanced equation to find the mole ratio between sodium and chlorine.
First, calculate the number of moles of sodium:
Mass of sodium = 4.81 g
Molar mass of sodium (Na) = 22.99 g/mol
Number of moles of sodium = Mass / Molar mass = 4.81 g / 22.99 g/mol = 0.2096 mol
According to the balanced equation, the mole ratio between sodium and chlorine is 1:1.
Therefore, the number of moles of chlorine gas required is also 0.2096 mol.
At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is equal to 22.4 L. So, we can calculate the volume of chlorine gas as:
Volume of chlorine gas = Number of moles of chlorine gas × Molar volume = 0.2096 mol × 22.4 L/mol = 4.688 L