1)
KOH + HBr = H2O +KBr
n(KOH) = 0.517*25.6/1000 = 0.0132 mol
n(KOH) = n(HBr) = 0.0132 mol
m( HBr) = 0.0132 * (1+81) = 1.0692 g
W( HBr) = 1.0692/14*100% = 7.64%
2)
2HNO3 + Ba(OH)2 = 2H2O + Ba(NO3)2
n(Ba(OH)2) = 1.18*18.7/1000 = 0.0221 mol
n(HNO3) = 2n(Ba(OH)2) = 0.0442 mol
m( HNO3) = 0.0442 * (1+14+48) = 1.39 g
W( HNO3) = 1.39/13.1*100% = 10.6%
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