how much iron(lll) chloride (in grams) is required to produce 2.33 liters and 1.24 molar iron(lll) chloride?
molarity = number of moles of iron(lll) chloride / volume of solution in L
1.24 = number of moles of iron(lll) chloride / 2.33
number of moles of iron(lll) chloride = 1.24 * 2.33 = 2.89 mole
number of moles of iron(lll) chloride = mass of iron(lll) chloride / molar mass of iron(lll) chloride
2.89 mole = mass of iron(lll) chloride / 162.2 g/mol
mass of iron(lll) chloride = 469 g
Therefore, the mass of iron(lll) chloride = 469 g
how much iron(lll) chloride (in grams) is required to produce 2.33 liters and 1.24 molar iron(lll) chloride?
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