Question

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1) How many moles of precipitate will be formed when 80.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ in the following chemical reaction? 2 NaI (aq) + Pb(NO₃)₂ (aq) → PbI₂ (s) + 2 NaNO₃ (aq)

Answer 1

Balanced Equator is ONO I ea) + p b c No3) caa) 2 role PhI2 (S) + 2Nanggar Iorde from Reactan 2 mae of Nal Reacts with lordle Pr No . to producu Inae pb Iz mole of Nat - mohoutex dume a = 2.0 molly Box 163 - 0.100 mole was mare ar pb.Ng) = = molarity of pb Nos) ₂x volume of pang o. quo mel poox le 32 = 0.09 mole phones) mae of pb No3) needed fg 0.100 male of Na I is lorde Pb (N? 0.160 met dat & - = 0.ogo mole 2 mode NOI Phng) we have Excess 1 phN3) i of Exess, pbus , 6.og male NaI is limiting reagent more of pbt from Nat is I oncle pbi2 O. 160 male Naax - 2 mde sal male e pot phity) = 0.080 mde pb.la