How many grams of AL(NO3)3 are produced when 23.5 grams of Aluminum react with 133 mL of 6.5M HCL?
To find the number of grams of AL(NO3)3 produced, we need to follow a few steps:
Step 1: Convert the volume of HCl from milliliters (mL) to liters (L).
To do this, divide the volume by 1000 since there are 1000 mL in 1 L:
133 mL ÷ 1000 = 0.133 L
Step 2: Calculate the number of moles of HCl.
To find the number of moles, we can use the formula: moles = concentration (M) × volume (L).
Given concentration: 6.5 M
Volume: 0.133 L
Using the formula: moles = 6.5 M × 0.133 L = 0.8645 moles
Step 3: Determine the balanced chemical equation between Aluminum (Al) and Hydrochloric acid (HCl) to find the mole ratio.
The balanced equation is: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
From the equation, we see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl3.
Step 4: Use the mole ratio to find the number of moles of AlCl3 produced.
Since the mole ratio is 2:2, which simplifies to 1:1, the number of moles of AlCl3 produced is also 0.8645 moles.
Step 5: Calculate the molar mass of AL(NO3)3.
The atomic mass of Aluminum (Al) is 26.98 g/mole.
The atomic mass of Nitrogen (N) is 14.01 g/mole.
The atomic mass of Oxygen (O) is 16.00 g/mole.
Since there are three oxygen atoms in AL(NO3)3, we multiply that atomic mass by 3 to get:
3 × 16.00 g/mole = 48.00 g/mole
Adding up the molar masses of each element, we get:
Al = 26.98 g/mole
N = 14.01 g/mole
O = 48.00 g/mole
Total molar mass of AL(NO3)3 = 26.98 g/mole + 3 × (14.01 g/mole + 48.00 g/mole)
= 26.98 g/mole + 3 × 62.01 g/mole
= 26.98 g/mole + 186.03 g/mole
= 213.01 g/mole
Step 6: Use the molar mass and moles of AlCl3 to calculate the grams of AL(NO3)3 produced.
Given moles of AlCl3 = 0.8645 moles
Molar mass of AL(NO3)3 = 213.01 g/mole
Using the formula: grams of AL(NO3)3 = moles of AlCl3 × molar mass of AL(NO3)3
Therefore, grams of AL(NO3)3 = 0.8645 moles × 213.01 g/mole = 184.06 grams
Therefore, 23.5 grams of Aluminum react with 133 mL of 6.5M HCl to produce 184.06 grams of AL(NO3)3.