CHEM 1040 Lecture Notes - Lecture 13: Aluminium Sulfate, Molar Mass, Chemical Equation
7 Nov 2017
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Chem 1040- lecture 13- titrations and stoichiometry review problems. Ex: when 22. 8ml of a 0. 150moll-1 kmno4soln are titrated with kno2 in the presence of acid, 5no3 (aq) + 2mn2+ (aq) + 3h2o(l) (aq) First, calculate the number of moles of kmno4: n= vc n= 0. 0228*0. 150 n= 0. 00342 mol. Next, calculate the moles of kno2 using the mole ratios: n= 0. 00342*(5/2) n= 0. 00855 mol. Calculate the concentration of kno2: c= n/v c= 0. 00855/0. 0322 c= 0. 2655mol/l. In an acid/base neutralization, strong acids are always used up completely. Ex: 500ml of 0. 250mol-1 hcl requires 250ml of a baoh2 solution to neutralize. You don"t need an equation for this question because you can use the ratio of the oh- and h+ in the acids and bases. Find the number of moles of hcl: n= vc n= 0. 500*0. 250 n= 0. 125mol. Find the number of moles of ba(oh)2 using mole ratios. n= 0. 125/2 n= 0. 0625mol.
