Asked by Pilly
5.5g of impure iron required 7.5cm3 of 3.0M HCL and give neutral solution calculate the purity of iron
Answers
Answered by
DrBob222
Fe + 2HCl ==> FeCl2 + H2
millimoles HCl = mL x M = 7.5 mL x 3 M = 22.5
Use the coefficients in the balanced equation to convert to millimoles Fe in the sample = 1/2 * 22.5 = 11.3 or 0.0113 moles.
grams Fe = mols x molar mass = 0.0113 x 55.85 = 0.628 g
Then % purity = (g Fe/g sample)*100 = (0.628/5.5)*100 = ?
millimoles HCl = mL x M = 7.5 mL x 3 M = 22.5
Use the coefficients in the balanced equation to convert to millimoles Fe in the sample = 1/2 * 22.5 = 11.3 or 0.0113 moles.
grams Fe = mols x molar mass = 0.0113 x 55.85 = 0.628 g
Then % purity = (g Fe/g sample)*100 = (0.628/5.5)*100 = ?
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