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  • A mixture of 56 g  and 40 g <img alt="\mathrm{NaOH}" src="https://entrancecorner.oncodec

A mixture of 56 g \mathrm{CaO} and 40 g \mathrm{NaOH} was made to react with aqueous \mathrm{HCl} will the reaction is complete drying the resulting solution: Calculate the mass of solid left behind.

Option: 1

169.5 g


Option: 2

84.75 g


Option: 3

42.37 g


Option: 4

100.0 g


Answers (1)

\begin{aligned} & \mathrm{CaO \; (s)}+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\ \\& \mathrm{NaOH}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \end{aligned}

As per the equations

Moles of \mathrm{CaCl_{2}=Moles\; of\; CaO\; =\frac{56}{56}=1}

Also, moles of \mathrm{NaCl=Moles\; of\; NaOH\; =\frac{40}{40}=1}

On evaporating the solution, only \mathrm{CaCl_{2}} and \mathrm{NaCl} are left in solid state.

Hence, the mass of the solid left =(1\times111)+(1\times 58.5)

=169.5 \; \text{g}

Posted by

Sumit Saini

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