Write plausible chemical equations for preparing each compound from the indicated starting material: (a) \(\operatorname{SnCl}_{2}\) from \(\operatorname{SnO} ;\) (b) \(\operatorname{SnCl}_{4}\) from \(\operatorname{Sn} ;\) (c) \(\operatorname{PbCrO}_{4}\) from \(\mathrm{PbO}_{2}\). What reagents (acids, bases, salts) and equipment commonly available in the laboratory are needed for each reaction?

Short Answer

Expert verified
The reactions, (a) \(\operatorname{SnO} + 2\operatorname{HCl} \rightarrow \operatorname{SnCl}_{2} + \operatorname{H}_{2}\operatorname{O}\), (b) \(\operatorname{Sn} + 4\operatorname{HCl} + catalyst \rightarrow \(\operatorname{SnCl}_{4} + 2\operatorname{H}_{2}\), and (c) \(\operatorname{PbO}_{2} + \operatorname{ K}_{2}\operatorname{CrO}_{4}\rightarrow \operatorname{PbCrO}_{4} + 2\operatorname{K}^{+} \operatorname{O}^{2-}\) show how each respective compound can be prepared from the starting material.

Step by step solution

01

Preparing SnCl2 from SnO

To obtain \(\operatorname{SnCl}_{2}\) from \(\operatorname{SnO}\), react \(\operatorname{SnO}\) with \(2\operatorname{HCl}\) where \(HCl\) is Hydrochloric acid. The reaction is: \(\operatorname{SnO} + 2\operatorname{HCl} \rightarrow \operatorname{SnCl}_{2} + \operatorname{H}_{2}\operatorname{O}\)
02

Preparing SnCl4 from Sn

To obtain \(\operatorname{SnCl}_{4}\) from \(\operatorname{Sn}\), react \(\operatorname{Sn}\) with \(4\operatorname{HCl}\). This reaction requires a catalyst to speed up: the Chlorination reaction. The reaction is: \(\operatorname{Sn} + 4\operatorname{HCl} \rightarrow \(\operatorname{SnCl}_{4} + 2\operatorname{H}_{2}\)
03

Preparing PbCrO4 from PbO2

Getting \(\operatorname{PbCrO}_{4}\) from \(\operatorname{PbO}_{2}\) involves adding \(\operatorname{PbO}_{2}\) to a solution of \( \operatorname{ K}_{2}\operatorname{CrO}_{4}\)(Potassium Chromate). The reaction is: \(\operatorname{PbO}_{2} + \operatorname{ K}_{2}\operatorname{CrO}_{4}\rightarrow \operatorname{PbCrO}_{4} + 2\operatorname{K}^{+} \operatorname{O}^{2-}\)

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