Determine the mass% of Fe^3+ in a 0.7450g sample of iron ore, if 22.40mL of a 0.1000M stannous chloride, SnCl2(aq) solution is required to completely react with the Fe^3+ present in the ore sample. The chemical equation for the reaction is 2Fe^3+(aq)+ Sn^2+(aq)-- 2Fe^2+(aq)+ Sn^4+(aq)Please show steps. Thank you.
Sounds bogus to me, unless you were doing a lot of deliberate skidding around, or the tires were noticeably underinflated.
moles of SnCl2 moles of Sn2+ M*V 0.1*22.4 2.24 millimoles Moles of Fe3+ 2*moles of Sn2+ 4.48 mmoles. Mass of Fe3+ 4.48*56 249.76 mg % of Fe3+ 0.24976/0.745*100% 33.53%
the moles of [c9e0e1712df7181d34e5cbe4ec1a643c9e0e171. Sn2+ C * V 0.a million * 0.0c9e0e1712df7181d34e5cbe4ec1a643c9e0e1. c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712. * 10^(-4) the moles of Fe 3c9e0e1712df7181d34e5cbe4ec1a643 that reacted c9e0e1712df7181d34e5cbe4ec1a643c9e0e1712. * Sn2+ / a million 40 4.8*10^(-4) the amouc9e0e1712df7181d34e5cbe4ec1a643t of Fe 3c9e0e1712df7181d34e5cbe4ec1a643 fifty six * 40 4.8 * 10^(-4) 0.c9e0e1712df7181d34e5cbe4ec1a6435088 g the mass percec9e0e1712df7181d34e5cbe4ec1a643t of Fe3c9e0e1712df7181d34e5cbe4ec1a643 0.c9e0e1712df7181d34e5cbe4ec1a6435088/0 * one hundred 38.89%
First figure out how many moles of SnCl2 you have: 22.40 mL (0.1000moles/L) (L/1000mL) 0.002240 moles Each mole of SnCl2 reacts with 2 moles of Fe^3+ (according to the equation) 2 x 0.002240 moles 0.004480 moles of Fe^3+ The mass of the Fe is 0.004480 moles x 55.85 g/mole 0.2502 g mass % (0.2502g/0.7450g) 100%33.58%
Something wrong with the bike , methinks.
