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?
5.0g = 5.0/26.98 = 0.185 moles Al
50 mL of 6M = 0.3 moles HCl
Since 2Al + 6HCl --> 2AlCl3 + 3H2, only 0.1 moles of Al is used, producing 0.1 moles AlCl3 and 0.15 moles H2.
so, the volume of gas is 0.15 * 22.4L
for the other problem, use PV=nRT to find the # moles. Then divide into the mass.
To solve these problems, we will use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
For problem 5.0 g of aluminum reacts with 50.0 mL of 6.0 M hydrochloric acid to produce hydrogen gas, we need to calculate the number of moles of aluminum and then use the stoichiometry of the reaction to determine the number of moles of hydrogen gas produced. Finally, we can use the ideal gas law to find the volume of gas.
1. Calculate the number of moles of aluminum:
Since the molar mass of aluminum (Al) is 26.98 g/mol, we can calculate the number of moles using the formula:
moles = mass / molar mass = 5.0 g / 26.98 g/mol = 0.185 mol
2. Determine the number of moles of hydrogen gas:
The balanced chemical equation for the reaction between aluminum and hydrochloric acid is:
2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)
From the equation, we can see that 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, the number of moles of hydrogen gas produced can be calculated as:
moles of H2 = (moles of Al) x (3 moles of H2 / 2 moles of Al) = 0.185 mol x (3/2) = 0.278 mol
3. Calculate the volume of the gas at STP:
At STP (standard temperature and pressure), the temperature (T) is 273.15 K, and the pressure (P) is 1 atm. The ideal gas constant (R) is 0.0821 L·atm/mol·K.
We can rearrange the ideal gas law equation to solve for volume (V) as follows:
V = (n x R x T) / P = (0.278 mol x 0.0821 L·atm/mol·K x 273.15 K) / 1 atm = 6.24 L
Therefore, the volume of gas collected, if the gas was collected at STP, is 6.24 L.
For problem 2, to find the molar mass of the gas, we can use the ideal gas law. Since we are given the volume, temperature, pressure, and mass of the gas, we need to calculate the number of moles and then use the formula for molar mass.
1. Calculate the number of moles of the gas:
Using the ideal gas law (PV = nRT), we can rearrange the equation to solve for moles (n):
n = PV / RT = (743 mmHg x 115 mL) / (0.0821 L·atm/mol·K x (66.3 °C + 273.15) K)
However, the pressure is given in mmHg and the volume is given in mL, so we need to convert them to atm and L, respectively:
1 atm = 760 mmHg
1 L = 1000 mL
Substituting the values and converting the units, we have:
n = (743 mmHg x 115 mL / 760 mmHg/atm x 1 atm) / (0.0821 L·atm/mol·K x (66.3 °C + 273.15) K) = 0.0042 mol
2. Calculate the molar mass of the gas:
The molar mass (M) of a substance is defined as the mass (m) of the substance divided by the number of moles (n):
M = m / n = 0.418 g / 0.0042 mol = 99.52 g/mol
Therefore, the molar mass of the gas is 99.52 g/mol.