5.0 g of aluminum reacts with 50.0 mL of 6.0 M hydrochloric acid to produce
hydrogen gas. What is the volume of gas collected if the gas was collected at STP?
3. A 0.418 g sample of gas has a volume of 115 mL at 66.3 °C and 743 mmHg. What is
the molar mass of this gas?
stop wasting everyone's time reposting the same thing over and over!
This is a limiting reagnt (LR) problem. You know that when amounts are given for BOTH reactants.
2Al + 6HCl ==>2 AlCl3 + 3H2
mols Al = grams/atomic mass
mols HCl = M x L = ?
Convert mols Al to mols H2 using the coefficients in the balanced equation. You can see that 2 mols Al = 3 mols H2.
Convert mols HCl to mols H2 .
In LR problems, the SMALLER number wins and that will be the mols H2 produced.
Volume is calculated using PV = nRT OR remembering that 1 mole of any gas occupies 22.4 L at STP.
For #3. use PV = nRT
Post your work if you get stuck.
To determine the volume of gas collected, we need to use the concept of stoichiometry and the ideal gas law.
For the first question, we have the following information:
Mass of aluminum (Al) = 5.0 g
Volume of hydrochloric acid (HCl) = 50.0 mL
Concentration of HCl = 6.0 M
Step 1: Convert the volume of HCl from milliliters (mL) to liters (L).
To do this, divide the volume by 1000:
50.0 mL ÷ 1000 = 0.050 L
Step 2: Use stoichiometry to determine the moles of HCl.
The balanced chemical equation for the reaction between Al and HCl is:
2Al + 6HCl → 3H2 + 2AlCl3
From the equation, we can see that 6 moles of HCl are required to produce 3 moles of H2.
Since the concentration of HCl is 6.0 M, we can calculate the moles of HCl:
0.050 L (volume of HCl) × 6.0 mol/L (concentration) = 0.30 mol HCl
Step 3: Use stoichiometry again to determine the moles of H2.
From the balanced chemical equation, we can see that 3 moles of H2 are produced from 6 moles of HCl.
So, the moles of H2 produced will also be half of the moles of HCl:
0.30 mol HCl ÷ 2 = 0.15 mol H2
Step 4: Use the ideal gas law to calculate the volume of H2 at STP.
The ideal gas law is given by PV = nRT, where:
P = pressure (in this case, at STP, pressure is approximately 1 atm)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in this case, at STP, temperature is 273 K)
Rearranging the equation to solve for V, we get:
V = nRT/P
Substituting the values into the equation:
V = (0.15 mol) × (0.0821 L·atm/mol·K) × (273 K) / (1 atm)
V ≈ 3.18 L
Therefore, if the gas was collected at STP, the volume of gas collected is approximately 3.18 liters.
For the second question, we have the following information:
Mass of the gas sample = 0.418 g
Volume of the gas = 115 mL
Temperature = 66.3 °C (convert to Kelvin by adding 273: 66.3 °C + 273 = 339.3 K)
Pressure = 743 mmHg (convert to atm by dividing by 760: 743 mmHg ÷ 760 = 0.976 atm)
Step 1: Convert the volume of the gas from milliliters (mL) to liters (L).
To do this, divide the volume by 1000:
115 mL ÷ 1000 = 0.115 L
Step 2: Use the ideal gas law to calculate the number of moles of the gas.
Rearranging the ideal gas law equation, we know that n (number of moles) is:
n = PV / RT
Substituting the values into the equation:
n = (0.976 atm) × (0.115 L) / [(0.0821 L·atm/mol·K) × (339.3 K)]
n ≈ 0.00397 mol
Step 3: Use the molar mass formula to calculate the molar mass of the gas.
Molar mass (g/mol) = mass (g) / moles
Substituting the values into the equation:
Molar mass = 0.418 g / 0.00397 mol
Molar mass ≈ 105.41 g/mol
Therefore, the molar mass of the gas is approximately 105.41 g/mol.