Duralin is an alloy of aluminium containing
4.5% manganese, 3.5% copper, and 0.45%
magnesium. How much magnesium is present
in a 32 g sample of this alloy?
Answer in units of mol
To find the amount of magnesium present in a 32 g sample of Duralin, we need to calculate the moles of magnesium using its molar mass.
The atomic mass of magnesium (Mg) is approximately 24.31 g/mol.
First, let's calculate the mass of magnesium in the 32 g sample. The percentage of magnesium in the alloy is 0.45%, which means there are 0.45 grams of magnesium for every 100 grams of alloy.
Mass of magnesium = (0.45/100) * 32 g
= 0.144 g
Now, let's convert the mass of magnesium to moles.
Moles of magnesium = Mass of magnesium / Molar mass of magnesium
= 0.144 g / 24.31 g/mol
= 0.00593 mol
Therefore, there is approximately 0.00593 moles of magnesium present in a 32 g sample of Duralin.
To calculate the amount of magnesium present in a 32 g sample of Duralin alloy, we first need to convert the mass of the alloy to moles.
To do this, we need to determine the molar mass of Duralin. The molar mass is simply the sum of the atomic masses of all the elements in the alloy.
The atomic mass of aluminum (Al) is 26.98 g/mol.
The atomic mass of manganese (Mn) is 54.94 g/mol.
The atomic mass of copper (Cu) is 63.55 g/mol.
The atomic mass of magnesium (Mg) is 24.31 g/mol.
Now, we can calculate the molar mass of Duralin:
Molar mass of Duralin = (mass of Al) + (mass of Mn) + (mass of Cu) + (mass of Mg)
= (4.5% of 26.98 g/mol) + (3.5% of 54.94 g/mol) + (0.45% of 63.55 g/mol) + (mass of Mg)
= 1.214 g + 1.922 g + 0.286 g + (mass of Mg)
Since the alloy weighs 32 g, we can set up the equation:
32 g = 1.214 g + 1.922 g + 0.286 g + (mass of Mg)
Now, let's solve for the mass of magnesium:
mass of Mg = 32 g - (1.214 g + 1.922 g + 0.286 g)
= 32 g - 3.422 g
= 28.578 g
Finally, we convert the mass of magnesium to moles by dividing by its molar mass:
moles of Mg = mass of Mg / molar mass of Mg
= 28.578 g / 24.31 g/mol
≈ 1.18 mol
Therefore, there is approximately 1.18 moles of magnesium present in a 32 g sample of Duralin alloy.
How many Mg WHAT?
g Mg = 32*0.0045 = ?
mols Mg = ?g Mg/atomic mass Mg = y
#atoms Mg = y mols Mg*6.02E23 = ??
32g*.45gMg/100gDuralin*1mol/24.305gMg
0.005924707mol Mg