Answer to Question #324926 in General Chemistry for Shadow

Question

A 0.2719g sample containing (blank)3 reacted with 20.00mL of 0.2254 M HCl. Given that HCl was excess. The excess HCl required exactly 20.00mL of 0.1041 M NaOH to reach the end-point using phenolphthalein indicator. Determine percentage purity of (blank)3 in the sample.

The reaction involved is

3( )+2 ( )→ 2( )+2 2O(l)

The titration reaction is

( )+NaOH(aq) Cl( )+ 2 (l)

Answer 1

by the titration reaction:

HCl +NaOH = NaCl + H₂O

1) n(HCl) = CV = 0.2254(20/1000) = 0.004508 mol = 4.508 * 10^-3 mol

2) n(NaOH) = CV = 0.1041(20/1000) = 0.002082 mol = 2.082 * 10^-3 mol

3) n(HCl_reacted) = n(HCl_initial) - n(HCl_excess) = 4.508 * 10^-3 mol - 2.082 * 10^-3 mol = 2.426 * 10^-3 mol

by the reaction involved:

3(blank3)+2HCl→ 2( )+2 2O(l)

4) n(blank):n(HCl) = 3 : 2

n(blank_reacted) = 3n(HCl)/2 = 3.639 * 10^-3 mol

5) m(blank_reacted) = n(blank_reacted) * M(blank) = 3.639 * 10^-3 mol * M(blank)

 6) Determine percentage purity of (blank) = (m(blank_reacted)/ m(blank_initial))*100% = (m(blank_reacted)/3.639 * 10^-3 mol)*100%

You need to know the name of the substance (the word (blank) in this task), calculate its molar mass (M(blank)), substitute in equation 5 and determine percentage purity of (blank) in equation 6