How many Liters of hydrogen gas are produced if 3.52 g of magnesium react with excess nitric acid at STP?
Mg + 2HNO3 = Mg(NO3)2 + H2
so, you get one mole of H2 for every mole of Mg
how many moles of Mg in 3.52g?
1 mole occupies 22.4L at STP, so ...
To determine the number of liters of hydrogen gas produced, we need to use the molar ratio of magnesium and hydrogen gas from the balanced chemical equation.
The balanced equation for the reaction of magnesium and nitric acid is:
Mg + 2HNO3 -> Mg(NO3)2 + H2
From the equation, we can see that 1 mole of magnesium will produce 1 mole of hydrogen gas.
First, we need to convert the mass of magnesium given (3.52 g) to moles using the molar mass of magnesium (24.31 g/mol):
Number of moles of magnesium = Mass of magnesium / Molar mass of magnesium
Number of moles of magnesium = 3.52 g / 24.31 g/mol
Number of moles of magnesium ≈ 0.145 moles (rounded to three decimal places)
Since 1 mole of magnesium produces 1 mole of hydrogen gas, the number of moles of hydrogen gas produced is also 0.145 moles.
Using the ideal gas law at STP (Standard Temperature and Pressure), we can convert moles of hydrogen gas to liters:
1 mole of gas at STP occupies 22.4 L
Number of liters of hydrogen gas = Number of moles of hydrogen gas × 22.4 L/mol
Number of liters of hydrogen gas ≈ 0.145 moles × 22.4 L/mol
Number of liters of hydrogen gas ≈ 3.248 L (rounded to three decimal places)
Therefore, approximately 3.248 liters of hydrogen gas will be produced.
To determine the volume of hydrogen gas produced when 3.52 g of magnesium reacts with excess nitric acid at STP (Standard Temperature and Pressure), we need to use the concept of stoichiometry.
1. First, we need the balanced chemical equation for the reaction between magnesium and nitric acid:
Mg + 2HNO₃ → Mg(NO₃)₂ + H₂
2. From the balanced equation, we can see that 1 mole of magnesium will produce 1 mole of hydrogen gas.
So, we need to convert the given mass of magnesium (3.52 g) to moles.
To do this, we need the molar mass of magnesium (Mg) from the periodic table, which is 24.31 g/mol.
Moles of magnesium = Mass of magnesium / Molar mass of magnesium
= 3.52 g / 24.31 g/mol
3. Now that we have the moles of magnesium, we can determine the number of moles of hydrogen gas produced, which is equal to the moles of magnesium.
4. Next, we can use the ideal gas law to convert the moles of hydrogen gas to volume at STP.
The ideal gas law equation is: PV = nRT
Where:
P = Pressure (which is 1 atm at STP)
V = Volume of gas (what we need to find)
n = Number of moles of gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (which is 273 K at STP)
Rearranging the equation, we have:
V = (n * R * T) / P
Substituting the values, we can calculate the volume of hydrogen gas.
Note: Since we are given the reaction occurs at STP, we can assume that the resulting hydrogen gas is at standard temperature and pressure.
Remember to double-check your calculations for accuracy!
I hope this explanation helps you to calculate the volume of hydrogen gas produced when 3.52 g of magnesium reacts with excess nitric acid at STP.