4826 to m4 aloh 2 ass 23 oles of naoh required ass

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Unformatted text preview: +x2 H2asslof Mg(OH) 3 HNO3(aq) + Al(OH)3(aq) → Mg(NO3(aq) ) 3 H2m O( ) O(l) et x = mass of Al(OH)3 and 3)2(aq = 2 + 2) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2 H2O(l) - 3+ Let x =+ H Crof( Al(OH)3 a C 0.4826 –to = m4 Al(OH) 2 ass 23 oles of NaOH required à༎ ass Mg(OH) 16H (aq) 2m2NOMaq (aq) + nd2H5OH(aq)x titrateCrof (aq)3+=2CO2(g) + 11H2O(l) O7 a Moles of x = mass of Al(OH)3 rate Al(OH)3 x = mass of Mg(OH)2 Let NaOH required to tit ) nd 0.4826 – = 4.71 a) +5 b) – c +5 = 0.038462 x Moles of NaOH required3to titrate Al(OH)3 = If 35.46 mL of 0.05961 M Cr2O72- is required to titrate 28.00 = 0.038462 x what is the g of plasma, 4.75 percent Mf alcohol iH+required to titrate Mg(OH)2 = = Sn R = 0.038462 x mass a) Oxidizing agent = n the blood? educing agent o oles of NaOH b) Oxidizing agent = HMg(OH)2 = Reducing agent = Fe2+ O2 Moles of NaOH required to titrate 2 = 0.01655 – 0.03429x Moles of NaOH required to titrate Mg(OH)2 = Mass C2H C2H5= 4.81 %Mass 5OH OH = = 0.01655 – 0.03429x Moles of HNO3 required to titrate sample = 0.0173 mol = 0.038462 = 0.01655 – 0.03429x x + (0.01655 oles of HNO required to titrate sample = 0.0173 mol = 0.038462 – 0.03429x) M 3 0.0173 = 0.004172x – 0.01655 x + (0.01655 – 0.03429x) Moles of HNO3 required to titrate sample = 0.0173 0.179770.038462 of Al(OH)3 mol = = x = mass x + (0.01655 – 0.03429x) = 0.0486907 g C2H5OH (unrounded) 0.0173 = 0.004172x – 0.01655 0.17977 Mass % of Al(OH)3 = = 37.2503 = 37.25% 0.0173 = = x = mass of0Al(OH)3 0.004172x – .01655 Mass percent C2H5OH = Mass % of Al(OH)3 = 4.64 4.64 = 0.173895 = 0.1739% C2H5OH 0.17977 = x==3mass of = 37.25% 7.2503 Al(OH)3 Mass % of Al(OH)3 = = 37.2503 = 37.25% or 4equivalent Ao oxidizingmass of Fean atomis above Uto lthe mass decreasesaduring the reaction.se) molar (v/v)% hich whose oxidation number o...
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