Using the reaction below, calculate how many grams of bromine gas are necessary to produce 1.00mL of hydrogen bromide gas at STP.
H`2 + Br`2 -> 2 HBr
where `2 is a subscript two.
One simple way to do this is calculate how many moles of HBr are in 1ml
Moles=.001L/22.4L
So, you have the number of moles of HBr. Since Br2 was used to make it, it must be 1/2 that number of moles. convert that to grams.
To calculate the amount of bromine gas required to produce 1.00 mL of hydrogen bromide gas at STP, we can use the concept of stoichiometry.
Step 1: Convert milliliters to liters
Given: 1.00 mL
1 mL = 0.001 L (since there are 1000 mL in 1 L)
Therefore, 1.00 mL = 0.001 L
Step 2: Determine the number of moles of hydrogen bromide
According to the balanced equation:
1 mole of H2 reacts with 1 mole of Br2 to produce 2 moles of HBr
So, the moles of HBr produced is equal to the moles of H2 and Br2 used.
Step 3: Use the Ideal Gas Law to determine the moles of bromine gas
The Ideal Gas Law is given by the formula:
PV = nRT
P = Pressure (STP is 1 atm)
V = Volume (0.001 L)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature in Kelvin (STP is 273.15 K)
Solving for n (moles):
n = PV / RT
Substituting the values:
n = (1 atm * 0.001 L) / (0.0821 L.atm/mol.K * 273.15 K)
Step 4: Since the balanced equation shows that 1 mole of Br2 reacts with 2 moles of HBr, we need to multiply the moles of HBr by 1/2 (0.5) to get the moles of Br2 required.
Step 5: Convert moles of Br2 to grams using its molar mass
The molar mass of Br2 is 159.808 g/mol.
Multiply the moles of Br2 by its molar mass to get the grams:
grams of Br2 = moles of Br2 * molar mass of Br2
Following these steps, you can calculate the grams of bromine gas required to produce 1.00 mL of hydrogen bromide gas at STP.
To calculate the grams of bromine gas needed to produce 1.00 mL of hydrogen bromide gas at STP, we need to follow these steps:
Step 1: Determine the molar ratio
From the balanced chemical equation, we can see that:
1 molecule of H2 reacts with 1 molecule of Br2 to produce 2 molecules of HBr.
This means that the molar ratio of H2 to Br2 is 1:1.
Step 2: Convert mL to moles of HBr
To calculate the number of moles of HBr produced, we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (STP, which is 1 atm)
V = volume (1.00 mL, which we need to convert to liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (273.15 K for STP)
Converting 1.00 mL to liters:
1.00 mL = 1.00 × 10^(-3) L
Substituting the values into the equation:
(1 atm) × (1.00 × 10^(-3) L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K)
Solving for n:
n = (1 atm × 1.00 × 10^(-3) L) / ((0.0821 L·atm/(mol·K)) × (273.15 K))
Step 3: Calculate the moles of Br2
Since the molar ratio of H2 to Br2 is 1:1, the moles of Br2 needed will be the same as the moles of HBr. So, the moles of Br2 is equal to n.
Step 4: Convert moles of Br2 to grams
To convert moles of Br2 to grams, we need to know the molar mass of Br2.
The molar mass of Br2 is:
1 mol Br = 79.90 g/mol
2 moles of Br = 2 × (79.90 g/mol) = 159.80 g/mol
Multiply the moles of Br2 by the molar mass to get the grams of Br2.
Finally, substitute the value of n (moles of Br2) into the equation:
Grams of Br2 = n × (159.80 g/mol)
This will give you the grams of bromine gas necessary to produce 1.00 mL of hydrogen bromide gas at STP.