A) \[3.4g\]
B) \[0.34g\]
C) \[0.34g\]
D) \[0.17g\]
Correct Answer: C
Solution :
\[\underset{(2+32=34)}{\mathop{BaC{{l}_{2}}+{{H}_{2}}S}}\,\xrightarrow{{}}\underset{(137+32=169)}{\mathop{BaS\downarrow +2HCl}}\,\] \[\because \] \[169g\text{ }BaS\]is obtained\[34g\]. \[{{H}_{2}}S\] \[\therefore \] \[1.69g\text{ }BaS\]will be obtained by \[=\frac{34\times 1.69}{169}\] \[=0.34g\,{{H}_{2}}S\]
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