Five grams of sodium are combined with unlimited bromine. How many moles of bromine are used up? How many liters of bromine would be used if the reaction took place at STP?
My answer:
.1088 mole Br2 and 2.44 L Br2
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.
To find out how many moles of bromine are used up in the reaction, we need to use stoichiometry, which involves using the balanced equation for the reaction. The balanced equation for the reaction between sodium and bromine is:
2 Na + Br2 → 2 NaBr
From the balanced equation, we can see that 1 mole of bromine reacts with 2 moles of sodium. Therefore, for every 2 moles of sodium, 1 mole of bromine is used.
Given that we have 5 grams of sodium, we can convert this to moles using the molar mass of sodium. The molar mass of sodium is 22.99 g/mol.
Moles of sodium = Mass of sodium / Molar mass of sodium
Moles of sodium = 5 g / 22.99 g/mol
Moles of sodium = 0.2175 mol
Using the mole ratio from the balanced equation, we can determine the moles of bromine that are used up in the reaction:
Moles of bromine = Moles of sodium / 2
Moles of bromine = 0.2175 mol / 2
Moles of bromine = 0.1088 mol
Therefore, 0.1088 moles of bromine are used up in the reaction.
To find out how many liters of bromine would be used if the reaction took place at STP (Standard Temperature and Pressure), we can use the ideal gas law. At STP, the pressure is 1 atm and the temperature is 273 K.
The ideal gas law is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
Rearranging the equation to solve for V:
V = (nRT) / P
Given that we have 0.1088 moles of bromine and using the values of R = 0.0821 L.atm/mol.K and P = 1 atm, we can find the volume:
V = (0.1088 mol * 0.0821 L.atm/mol.K * 273 K) / 1 atm
V = 2.44 L
Therefore, if the reaction took place at STP, approximately 2.44 liters of bromine would be used.