How to solve this mole problem (chemistry)?
"how many moles of chloride ions are in 0.0750 g of AlCl3 (aluminum chloride)?"
TY.
Answer:
0.0750g AlCl3 * ( 1 mol AlCl3 / 133.341 g AlCl3 ) = 0.000562 mol AlCl3
Now, you know that there is one mole of Al and 3 moles of Cl in every mole of AlCl3.
This means that there are 3 * 0.000562 moles of Cl, or 0.00169 mol of Cl- in 0.0750g of AlCl3
Hope this helps!
AlCl3 Atomic mass = 27 + (3 x 35.5) = 133.5g/mole.
Cl3 = 106.5g/mol
106.5 ÷ 133.5 = 0.798g/mol.Cl3
0.0750g ÷ 0.0.798g/mol = 0.094 moles Cl3.
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TY.
Answer:
0.0750g AlCl3 * ( 1 mol AlCl3 / 133.341 g AlCl3 ) = 0.000562 mol AlCl3
Now, you know that there is one mole of Al and 3 moles of Cl in every mole of AlCl3.
This means that there are 3 * 0.000562 moles of Cl, or 0.00169 mol of Cl- in 0.0750g of AlCl3
Hope this helps!
AlCl3 Atomic mass = 27 + (3 x 35.5) = 133.5g/mole.
Cl3 = 106.5g/mol
106.5 ÷ 133.5 = 0.798g/mol.Cl3
0.0750g ÷ 0.0.798g/mol = 0.094 moles Cl3.
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