Problems A
423
Vol of H 2 0 required = (4.7 g NaCl)
Q =
13 -
2 ml H
2°
13.2 ml - 11.9 ml = 1.3 ml water added to the filtrate
Wt of KN0 3 dissolved in 13.2 ml = (13.2 ml H 2 O) ( '"
g ™°
3 } = 1.8gKNO 3
\ 1UU ml n.2U /
The undissolved KNO 3 (29.4 g - 1.8 g = 27.6 g pure KNO 3 ) is filtered off.
At the end of two cycles of fractional crystallization, you have
421.0 + 74.3 = 495.3 g pure NaCl
470.6 + 27.6 = 498.2 g pure KNO 3
In practice, you would have to use volumes of water slightly greater than the
calculated amounts to avoid inevitable contamination of one salt by the other; the
yields of pure salts therefore would not be quite as high. Great care must be taken
to keep the solutions at 0.0°C and 100. 0°C during the filtrations.
PROBLEMS
1. None of the reactions listed below takes place. Explain why.
(a) 2KNO 3 + ZnBr 2 -» 2KBr + Zn(NO 3 ) 2
(b) NaCl + H 2 O -» NaOH + HC1
(c) Br 2 + 2NaCl -> 2NaBr + C1 2
(d) CaCO 3 + 2NaCl -» CaCl 2 + NajCOa
(e) 2Au + 2H 3 PO 4 -> 2AuPO 4 + 3H 2
(f) KMnO 4 + 5Fe(NO 3 ) 3 + 8HNO 3
-> Mn(NO 3 ) 2 + 5Fe(NO 3 ) 4 + KNO 3 + 4H 2 O
(g) H 2 S + MgCl 2 -» MgS + 2HC1
2. The following reactions go to the right. Rewrite each in ionic form, and tell
why it goes to the right. An excess of the second reactant is used.
(a) Mg(OH) 2 + 2NH 4 C1 -> MgCl 2 + 2NH 3 + 2H 2 O
(b) AgCl + 2NH 3 -» Ag(NH 3 ) 2 Cl
(c) Ag 2 S + 4KCN -> K 2 S + 2KAg(CN) 2
(d) AgCl + KI -> Agl + KC1
(e) SrSO 4 + Na.jCO 3 -> SrCO 3 + Na2SO 4
(f) FeSO 4 + H 2 S + 2NaOH -> FeS + NajSO, + 2H 2 O
(g) NH 4 C1 + NaOH -» NH 3 + H 2 O + NaCl
(h) KBr + H 3 PO 4 (conc.) -> HBr + KH 2 PO 4
(i) 3CuS + 8HNO 3 -> 3Cu(NO 3 ) 2 + 3S + 2ND + 4H 2 O
(j) Ag(NH 3 ) 2 Cl + KI -+ Agl + KC1 + 2NH 3
3. Give balanced ionic equations for the following reactions in aqueous solution.
If a reaction does not occur, write NR. Indicate precipitates by ], and gases
by f.
423
Vol of H 2 0 required = (4.7 g NaCl)
Q =
13 -
2 ml H
2°
13.2 ml - 11.9 ml = 1.3 ml water added to the filtrate
Wt of KN0 3 dissolved in 13.2 ml = (13.2 ml H 2 O) ( '"
g ™°
3 } = 1.8gKNO 3
\ 1UU ml n.2U /
The undissolved KNO 3 (29.4 g - 1.8 g = 27.6 g pure KNO 3 ) is filtered off.
At the end of two cycles of fractional crystallization, you have
421.0 + 74.3 = 495.3 g pure NaCl
470.6 + 27.6 = 498.2 g pure KNO 3
In practice, you would have to use volumes of water slightly greater than the
calculated amounts to avoid inevitable contamination of one salt by the other; the
yields of pure salts therefore would not be quite as high. Great care must be taken
to keep the solutions at 0.0°C and 100. 0°C during the filtrations.
PROBLEMS
1. None of the reactions listed below takes place. Explain why.
(a) 2KNO 3 + ZnBr 2 -» 2KBr + Zn(NO 3 ) 2
(b) NaCl + H 2 O -» NaOH + HC1
(c) Br 2 + 2NaCl -> 2NaBr + C1 2
(d) CaCO 3 + 2NaCl -» CaCl 2 + NajCOa
(e) 2Au + 2H 3 PO 4 -> 2AuPO 4 + 3H 2
(f) KMnO 4 + 5Fe(NO 3 ) 3 + 8HNO 3
-> Mn(NO 3 ) 2 + 5Fe(NO 3 ) 4 + KNO 3 + 4H 2 O
(g) H 2 S + MgCl 2 -» MgS + 2HC1
2. The following reactions go to the right. Rewrite each in ionic form, and tell
why it goes to the right. An excess of the second reactant is used.
(a) Mg(OH) 2 + 2NH 4 C1 -> MgCl 2 + 2NH 3 + 2H 2 O
(b) AgCl + 2NH 3 -» Ag(NH 3 ) 2 Cl
(c) Ag 2 S + 4KCN -> K 2 S + 2KAg(CN) 2
(d) AgCl + KI -> Agl + KC1
(e) SrSO 4 + Na.jCO 3 -> SrCO 3 + Na2SO 4
(f) FeSO 4 + H 2 S + 2NaOH -> FeS + NajSO, + 2H 2 O
(g) NH 4 C1 + NaOH -» NH 3 + H 2 O + NaCl
(h) KBr + H 3 PO 4 (conc.) -> HBr + KH 2 PO 4
(i) 3CuS + 8HNO 3 -> 3Cu(NO 3 ) 2 + 3S + 2ND + 4H 2 O
(j) Ag(NH 3 ) 2 Cl + KI -+ Agl + KC1 + 2NH 3
3. Give balanced ionic equations for the following reactions in aqueous solution.
If a reaction does not occur, write NR. Indicate precipitates by ], and gases
by f.