硫酸亚锡(SnSO 4 )是一种重要的硫酸盐,主要用于电镀工业的镀锡、铝合金表面的氧化着色、印染工业的媒

2025-03-18 01:10:47
推荐回答(1个)
回答1:

SnCl 2 粉末加浓盐酸进行溶解得到酸性溶液,此时溶液中含有Sn 2+ 、Sn 4+ ,向其中加入Sn粉,Sn粉可以和H + 发生反应,使溶液酸性减弱,调节了溶液pH值,另外Sn可以将被氧化生成的Sn 4+ 还原成Sn 2+ ,即防止Sn 2+ 被氧化为Sn 4+ ,过滤得SnCl 2 溶液,向其中加碳酸钠,将Sn元素以SnO形式沉淀,过滤洗涤得纯净的SnO,加稀硫酸,得SnSO 4 溶液,加热浓缩、冷却结晶、过滤、洗涤,制得SnSO 4 晶体;
(1)SnCl 2 在水中发生水解反应:SnCl 2 +H 2 O═Sn(OH)Cl+HCl(可逆反应,应该用可逆号),生成难溶物Sn(OH)Cl,溶液含有杂质,若加入HCl,可使平衡向逆反应方向移动,抑制Sn 2+ 水解;
故答案为:加入盐酸,使水解平衡向左移动,抑制Sn 2+ 水解;
(2)由于在酸性条件下,锡在水溶液中有Sn 2+ 、Sn 4+ 两种主要存在形式,Sn 2+ 易被氧化,加入Sn粉,Sn粉可以和H + 发生反应,使溶液酸性减弱,调节了溶液pH,另外Sn可以将被氧化生成的Sn 4+ 还原成Sn 2+ ,即防止Sn 2+ 被氧化为Sn 4+
故答案为:防止Sn 2+ 被氧化;
(3)反应Ⅰ为向SnCl 2 溶液中加碳酸钠,得到的沉淀为SnO,该沉淀经过滤后,表面附着着Cl - ,要想检验滤渣是否洗涤干净,只要检验最后一次洗涤液中是否含有Cl - ,若没有,则证明已经洗涤干净,否则没有洗涤干净,用AgNO 3 溶液检验Cl - ,方法:取最后一次洗涤液,向其中加入AgNO 3 溶液,若无沉淀,则说明已洗涤干净,
故答案为:取最后一次洗涤液,向其中加入AgNO 3 溶液,若无沉淀,则说明已洗涤干净;
(4)反应Ⅱ已经得到SnSO 4 溶液,由溶液得到晶体的方法、步骤为:加热浓缩、冷却结晶、过滤、洗涤,
故答案为:加热浓缩、冷却结晶、过滤、洗涤;
(5)酸性条件下,SnSO 4 还可以用作双氧水去除剂,即SnSO 4 在酸性条件下和双氧水反应,Sn 2+ 有还原性,双氧水有氧化性,产物应该是:Sn 4+ 和H 2 O,方程式为:Sn 2+ +H 2 O 2 +2H + =Sn 4+ +2H 2 O,
故答案为:Sn 2+ +H 2 O 2 +2H + =Sn 4+ +2H 2 O;
(6)①②③发生的反应分别为Sn+2HCl→SnCl 2 +H 2 ↑①,SnCl 2 +2FeCl 3 =SnCl 4 +2FeCl 2 ②,6FeCl 2 +K 2 Cr 2 O 7 +14HCl→6FeCl 3 +2KCl+2CrCl 3 +7H 2 O6Sn~K 2 Cr 2 O 7 ③由方程式①②③得知K 2 Cr 2 O 7 ~6FeCl 2 ~3SnCl 2 ~3Sn,
n(Sn)=3n(K 2 Cr 2 O 7 )=3×0.1000mol/L×0.032L=0.0096mol,m(Sn)=n(Sn)×M(Sn)=0.0096mol×119g/mol=1.1424g,
锡粉样品中锡的质量分数=
m(Sn)
m(样品)
×100%=
1.1424g
1.226g
×100%=93.18%,
故答案为:93.18%.

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