To determine the moles of bromine used up, we need to first calculate the moles of sodium used:
1. Calculate the moles of sodium:
Given: Mass of sodium = 5 grams.
The molar mass of sodium ( Na ) is 23 g/mol.
Moles of sodium (Na) = Mass of sodium / Molar mass of sodium = 5 g / 23 g/mol = 0.2174 mol Na.
Now, we need to determine the moles of bromine used in the reaction. Since there is excess bromine, it is the limiting reagent.
2. Calculate the moles of bromine used:
The balanced chemical equation is:
2 Na + Br2 -> 2 NaBr
From the equation, we can see that two moles of sodium react with one mole of bromine to form two moles of sodium bromide.
Since we have 0.2174 moles of sodium, the moles of bromine used will be half of that value, as the ratio is 2:1.
Moles of bromine used = 0.2174 mol Na / 2 = 0.1087 mol Br2 (rounded to four decimal places).
To find the volume of bromine used at STP, we need to use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature.
As the reaction took place at STP, we can use the following values:
- Pressure (P) = 1 atm
- Gas constant (R) = 0.0821 L·atm/(mol·K)
- Temperature (T) = 273 K
3. Calculate the volume of bromine used at STP:
From the balanced equation, we know that 1 mole of a gas occupies 22.4 liters at STP.
Using the mole-to-volume conversion, we can find the volume of bromine gas used:
Volume (V) = moles of bromine (n) * 22.4 L/mol
Volume of bromine used = 0.1087 mol Br2 * 22.4 L/mol = 2.435 L Br2 (rounded to three decimal places).
Therefore, the moles of bromine used up in the reaction would be approximately 0.1087 mol, and the volume of bromine used at STP would be approximately 2.435 liters.