what is the mass percent of Fe3+ in the sample of a .6450g of iron ore requires 22.40mL of .1000M Sn2+ to completely react with the Fe3+ content of the ore?
1. 19.40%
2. 25.02%
3. 38.79%
4. 77.59%
2006-10-30 06:03:01 · 2 answers · asked by Shawn 1 in Science & Mathematics ➔ Chemistry
This is just a stoichiometry problem. First you can calculate how many moles of SN2+ were required to react with the iron in the sample (Molarity X volume = number of moles). Then from the balanced equation, you can calculate how many moles of iron were in the sample. Then, you can convert that to grams of iron with the molar mass. That answer is the mass of iron in the original 0.6450 gram sample. Divide the mass of iron by the mass of the sample and multiplly to 100 to make it a percent, and you're there....
2006-10-30 06:09:14 · answer #1 · answered by hcbiochem 7 · 1⤊ 0⤋
the moles of [c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a643] Sn2+ = C * V = 0.one million * 0.0c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a64340 = c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a643.4 * 10^(-4) the moles of Fe 3c9e0e1712df7181d34e5cbe4ec1a643 that reacted c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712df7181d34e5cbe4ec1a643.4*10^(-4) * Sn2+ / one million= forty 4.8*10^(-4) the amouc9e0e1712df7181d34e5cbe4ec1a643t of Fe 3c9e0e1712df7181d34e5cbe4ec1a643 fifty six * forty 4.8 * 10^(-4) = 0.c9e0e1712df7181d34e5cbe4ec1a6435088 g the mass percec9e0e1712df7181d34e5cbe4ec1a643t of Fe3c9e0e1712df7181d34e5cbe4ec1a643 0.c9e0e1712df7181d34e5cbe4ec1a6435088/0.6450 * a hundred = 38.89%
2016-12-16 16:38:04 · answer #2 · answered by ? 3 · 0⤊ 0⤋