Ten grams of Aluminum reacts with Sulfuric Acid to Form Aluminum Sulfate
a) How many grams of aluminum sulfate are formed?
A) 113 grams of aluminum sulfate are formed.
b) How many liters of hydrogen are obtained under standard conditions?
b) 28.1 liters of hydrogen gas are obtained under standard conditions.
Answer:
To calculate the grams of aluminum sulfate formed, we need to know the balanced chemical equation for the reaction between aluminum and sulfuric acid.
a) Calculation of Aluminum Sulfate:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
From this equation, we can see that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. The molar mass of aluminum sulfate is 342.15 g/mol, so we can calculate:
9 g Al × (1 mol Al/26.98 g Al) × (1 mol Al₂(SO₄)₃/2 mol Al) × (342.15 g Al₂(SO₄)₃/mol Al₂(SO₄)₃) = 113 g Al₂(SO₄)₃
b) Calculation of Hydrogen Gas:
Under standard conditions (0°C and 1 atm pressure), 1 mole of gas occupies 22.4 L of volume. From the balanced chemical equation above, we can see that 3 moles of hydrogen gas are produced for every 2 moles of aluminum that react. Therefore:
9 g Al × (1 mol Al/26.98 g Al) × (3 mol H2/2 mol Al) = 1.252 mol H2
PV = nRT
(1 atm)(V) = (1.252 mol)(0.08206 L·atm/mol·K)(273 K)
V = 28.1 L
Aluminum is a versatile metal used in various industries due to its unique properties such as high strength-to-weight ratio and corrosion resistance.
Aluminum Sulfate Formation:
When aluminum reacts with sulfuric acid, it forms aluminum sulfate and hydrogen gas. The balanced chemical equation allows us to calculate the amount of aluminum sulfate formed.
Hydrogen Gas Production:
Using the ideal gas law, the volume of hydrogen gas produced can be determined under standard conditions of temperature and pressure.
Conclusion:By understanding the chemical reactions involving aluminum and sulfuric acid, we can calculate the quantities of aluminum sulfate and hydrogen gas produced in the given scenario.