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Amount of Substance (includes back titration)
01/04/2023
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moles = mass mr moles = vol. x conc. moles vol. 24 m mr h m mr n K 56.5 с 8.7 12 m mr n To find emperical formula → simplest whole number ratio of atoms of elements in a compound. 39.1 1.445 0.725 1.993: I : 2 : : I K₂CO3 gases at rpt MgSO4. x H ₂0 water of crystallisation. 0 34.8 16 2.175 3 C₂ H₂ +50₂ 4.4 44 0.1 3 RTP temp = 25°C (298k) pressure = 101 kpa % yield = actual amount x 100 theoretical amount atom economy 120.4 0.406 mol PV = nRT → MgSO4 + x H₂O 48.9 2 often % composition in a = by lowest more → find x 3 CO₂ + 4H₂O constant 51.1 18 2.838 = ideau gas law mr desired product x 100 mr of reactants 0.3 0.3x24 = 7.2 dm³ add 273 to convert °C to K MgSO4 H₂O 0.406 2.838 : 6.989 : I 1:7 :: x = 7 if given volume, you don't need to work out mores Back Titrations Crock mainly сасоз mass of са соз mol v01 Conc Na O H 1.6 x 10-3 0.016 0.1 moles of acid that reacted excess acid YO. M mr 100.1 n N excess-leftover 50cm ³ of 0.5m HCl added with 1.39 IMPURE CaCO3 Solution up to 250cm³ w/ water. 25cm³ titrated against 0.1 m NaOH, took 1.6cm ³ + HICL - 1.6x10-3 0.025 mor of HCI in 25 cm³ = 1.6×10-3 44 "1 in 250 cm³ = 1.6 × 10-² 4.5X10-39x10-3 4.5×10/2 original acid : mol = voi x conc Solution of leftover acid 250cm³ moles of NaCl + H₂0 Teftover acid percentage purity of CaCO3 similar to to yield X10 FIND MOT of ACID INTO 250 CM³ = 0.05 x 0.5 = 0.025 mor → CaCl₂ + CO₂ + H₂O find moles of acid acid that REACTED with CaCO3 = 0.025 -1.6x10-² =9x10-3 mol саСО3 + 2на 0.450459 Jxn 0.016 dm² alkali, known conc. 0.45045 x 100 = 34.65% 1.3 25cm3...
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Alternative transcript:
of leftover acid