If you wanted to prepare 50.0 mL of hydrogen, collected over water at 25 C on a day when the barometric pressure was 730 torr, what mass of aluminum would you react with a hydrochloric acid? (Balance the equation first.)
___Al(s)+___HCl(aq)=___AlCl3(aq)+___H2(g)
To determine how much aluminum is needed to produce 50.0 mL of hydrogen gas over water, we first need to balance the chemical equation:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.
Next, we need to calculate the volume of hydrogen gas at 25 °C and 730 torr (the partial pressure of hydrogen). To do this, we can use the Ideal Gas Law:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹)
T = temperature (in Kelvin)
Converting the given values:
Pressure = 730 torr = 730/760 atm ≈ 0.9618 atm
Volume = 50.0 mL = 50.0/1000 L = 0.05 L
Temperature = 25 °C = 25 + 273 = 298 K
Plugging the values into the Ideal Gas Law equation, we can solve for the number of moles of hydrogen:
0.9618 atm x 0.05 L = n x 0.0821 L·atm·K⁻¹·mol⁻¹ x 298 K
0.04809 = n x 24.44218
n ≈ 0.001964 moles
Since the balanced equation tells us that 2 moles of aluminum produce 3 moles of hydrogen, we can calculate the number of moles of aluminum needed:
n(Al) = (2 moles Al / 3 moles H2) x n(H2)
n(Al) = (2/3) x 0.001964 ≈ 0.001309 moles Al
Finally, we can calculate the mass of aluminum needed using its molar mass, which is 26.98 g/mol:
Mass(Al) = n(Al) x molar mass(Al)
Mass(Al) = 0.001309 moles x 26.98 g/mol ≈ 0.03529 g
Therefore, you would need approximately 0.03529 grams of aluminum to react with hydrochloric acid and produce 50.0 mL of hydrogen gas collected over water at 25 °C, with a barometric pressure of 730 torr.
To determine the mass of aluminum needed to produce 50.0 mL of hydrogen gas, we need to first balance the equation and then use stoichiometry.
The balanced equation for the reaction is:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.
To find the moles of hydrogen gas produced, we can use the ideal gas law:
PV = nRT
Where:
P = pressure of the gas in atm (converted from torr)
V = volume of the gas in liters (converted from mL)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature of the gas in Kelvin (converted from degrees Celsius)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15 = 298.15 K
Next, let's convert the volume from mL to L:
V(L) = V(mL) / 1000
V(L) = 50.0 mL / 1000 = 0.050 L
Now, let's convert the pressure from torr to atm:
P(atm) = P(torr) / 760
P(atm) = 730 torr / 760 = 0.9618 atm
Substituting these values into the ideal gas law equation:
(0.9618 atm)(0.050 L) = n(0.0821 L·atm/mol·K)(298.15 K)
Solving for n (moles of hydrogen gas):
n = (0.9618 atm * 0.050 L) / (0.0821 L·atm/mol·K * 298.15 K)
n = 0.0305 moles
From the balanced equation, we know that 3 moles of hydrogen gas are produced from 2 moles of aluminum. Therefore, the moles of aluminum needed can be determined using a ratio:
(2 moles Al / 3 moles H2) = (x moles Al / 0.0305 moles H2)
Solving for x:
x = (2 moles Al)(0.0305 moles H2) / 3 moles H2
x = 0.0202 moles Al
To convert moles of aluminum to grams, we need to multiply by the molar mass of aluminum, which is 26.98 g/mol:
Mass of aluminum = (0.0202 moles Al)(26.98 g/mol) = 0.545 g
Therefore, you would need approximately 0.545 grams of aluminum to react with hydrochloric acid and produce 50.0 mL of hydrogen gas collected over water at 25 °C on a day when the barometric pressure was 730 torr.