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Mixture $$X$$ containing $$0.02$$ mol of $$[Co(NH_3)_5SO_4]Br$$ and $$0.02$$ mol of $$[Co(NH_3)_5Br]SO_4$$ was prepared in $$2$$ L solution.
$$1$$ L of mixture $$X$$ + excess of $$AgNO_3$$ solution $$\rightarrow Y$$
$$1$$ L of mixture $$X$$ + excess of $$BaCl_2$$ solution $$\rightarrow Z$$
Number of moles of $$Y$$ and $$Z$$ is, respectively :


A
0.01,0.01
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B
0.02,0.01
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C
0.01,0.02
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D
0.02,0.02
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Solution

The correct option is A $$0.01,\: 0.01$$
$$1$$ L of mixture $$X$$ contains $$\dfrac {1} {2} \times 0.02 =0.01$$  moles of $$[Co(NH_3)_5SO_4]Br$$.
It will react with excess of silver nitrate to produce $$0.01$$ mole of $$AgBr$$ which is compound $$Y$$.
The reaction is as follows:
$$[Co(NH_3)_5SO_4]Br+AgNO_3\rightarrow [Co(NH_3)_5SO_4]NO_3+AgBr$$
$$1$$ L of mixture $$Y$$ contains $$\dfrac {1} {2} \times 0.02 =0.01$$  moles of $$[Co(NH_3)_5Br]SO_4$$.
It will react with excess of barium chloride to produce $$0.01$$ mole of barium sulfate which is compound $$Z$$.
The reaction is as follows:
$$[Co(NH_3)_5Br]SO_4+BaCl_2\rightarrow [Co(NH_3)_5Br]Cl_2+BaSO_4$$

Chemistry

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