When 2.05 grams of aluminum metal are dropped in 650. mL of 6.0 M hydrochloric acid, how many liters of gas are evolved at 20°C, 1.07 atm?
To find the number of liters of gas evolved in this reaction, we need to use stoichiometry and the ideal gas law.
First, we need to balance the chemical equation for the reaction between aluminum and hydrochloric acid:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
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 number of moles of aluminum:
moles of aluminum = mass of aluminum / molar mass of aluminum
The molar mass of aluminum is 26.98 g/mol. So,
moles of aluminum = 2.05 g / 26.98 g/mol = 0.076 moles
Since the stoichiometry of the reaction tells us that 2 moles of aluminum react to produce 3 moles of hydrogen gas, we can calculate the number of moles of hydrogen gas produced:
moles of hydrogen gas = (0.076 moles aluminum) x (3 moles H2 / 2 moles Al) = 0.114 moles
Now, to find the volume of gas evolved at STP (Standard Temperature and Pressure, which is 0°C or 273 K and 1 atm), we can use the ideal gas law:
PV = nRT
Here, 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.
We are given the pressure (1.07 atm) and the temperature (20°C, which we need to convert to K):
T(K) = T(°C) + 273 = 20°C + 273 = 293 K
Now, we can rearrange the ideal gas law to solve for volume:
V = nRT / P
Using the given values, we have:
V = (0.114 moles) x (0.0821 L·atm/(mol·K)) x (293 K) / (1.07 atm)
Calculating this expression, we find that V ≈ 0.276 L. Therefore, approximately 0.276 liters of gas are evolved at 20°C and 1.07 atm.