If 17.14 grams of nitric acid are reacted with a large amount of aluminum metal, what volume of of hydrogen gas would be produced at STP?
What would be the balanced equation for this problem?
3Al + 6HNO3 ==> 3Al(NO3)2 + 3H2
To determine the balanced equation for this reaction, we first need to know the chemical formula of nitric acid and aluminum. The formula for nitric acid is HNO3, while aluminum is represented as Al.
The balanced chemical equation for the reaction between nitric acid and aluminum can be written as follows:
3 HNO3 + Al -> Al(NO3)3 + 3 H2
In this equation, 3 moles of nitric acid react with 1 mole of aluminum to produce 1 mole of aluminum nitrate and 3 moles of hydrogen gas.
Now, to determine the volume of hydrogen gas produced at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation: PV = nRT.
At STP, the standard temperature is 273.15 K (0°C) and the standard pressure is 1 atm.
To find the volume of hydrogen gas, we need to calculate the number of moles of hydrogen gas produced. We can use stoichiometry to convert the known mass of nitric acid to moles of hydrogen gas.
1 mole of H2 = 3 moles of HNO3
First, convert the given mass of nitric acid (17.14 grams) to moles using its molar mass. The molar mass of HNO3 is 63.01 g/mol.
moles of HNO3 = (mass of HNO3 / molar mass of HNO3)
= (17.14 g / 63.01 g/mol)
= 0.2721 mol
Since the stoichiometric ratio is 1:3 for HNO3 to H2, we can determine the moles of H2 produced.
moles of H2 = 3 * moles of HNO3
= 3 * 0.2721 mol
= 0.8163 mol
Now, we can use the ideal gas law to calculate the volume of hydrogen gas produced.
V = (n * R * T) / P
Where:
V = volume of gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 atm/mol·K)
T = temperature (in Kelvin)
P = pressure (in atm)
Plugging in the values:
V = (0.8163 mol * 0.0821 atm/mol·K * 273.15 K) / 1 atm
V = 19.53 L
Therefore, approximately 19.53 liters of hydrogen gas would be produced at STP.