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ICSE Class 10 Chemistry 2019 Paper

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Copper sulphate solution reacts with sodium hydroxide solution to form a precipitate of copper hydroxide according to the equation :
2NaOH+CuSO4→Na2SO4+Cu(OH)2↓2\mathrm{NaOH}+\mathrm{CuSO}_4\rightarrow\mathrm{Na}_2\mathrm{SO}_4+\mathrm{Cu}(\mathrm{OH})_2\downarrow
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Question : 79 of 86
Marks: +1, -0
What mass of copper hydroxide is precipitated by using 200 gm200\ \mathrm{gm} of sodium hydroxide?
[H=1, O=16, Na=23, S=32, Cu=64   ]   [ H=1,\ O=16,\ Na=23,\ S=32,\ Cu=64 \;\text{ ] }\;
Solution:  
2NaOH+CuSO4→Na2SO4+Cu(OH)2↓2[23+16+1]64+16×2+2×1=2[40]  =64+32+2=80  =98\begin{array}{lcl} 2\mathrm{NaOH}+\mathrm{CuSO}_4 & \rightarrow & \mathrm{Na}_2\mathrm{SO}_4+\mathrm{Cu}(\mathrm{OH})_2 \downarrow \\ 2[23+16+1] & & 64+16\times 2+2\times 1 \\ =2[40] & & \; =64+32+2 \\ =80 & \; & =98 \end{array}
If 80 gm NaOH80\ \mathrm{gm}\ \mathrm{NaOH} produces
=98 gm   of   Cu(OH)2=98\ \mathrm{gm}\;\text{ of }\;\mathrm{Cu}(\mathrm{OH})_2
then 200 gm NaOH200\ \mathrm{gm}\ \mathrm{NaOH} produces
  =  9880×200\;=\;\frac{98}{80} \times 200
  =245 gm\;=245\ \mathrm{gm}
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