Question and Answers Forum

All Questions      Topic List

Solutions Questions

Previous in All Question      Next in All Question      

Previous in Solutions      Next in Solutions      

Question Number 178166 by Spillover last updated on 13/Oct/22

$$8.5\mathrm{g}\mathrm{of}\mathrm{hydrated}\mathrm{copper}\left(\mathrm{ii}\right)$$$$\mathrm{sulphate}{\mathrm{CuSO}}_4.{\mathrm{xH}}_2\mathrm{O}\mathrm{was}\mathrm{heated}$$$$\mathrm{to}\mathrm{dryness}.\mathrm{if}4.0\mathrm{g}\mathrm{of}\mathrm{anhydrous}$$$$\mathrm{copper}\left(\mathrm{ii}\right)\mathrm{sulphate}\left[{\mathrm{CuSO}}_4\right]\mathrm{obtain}$$$$\mathrm{Find}\mathrm{number}\mathrm{of}\mathrm{molecule}\mathrm{of}\mathrm{water}\mathrm{of}$$$$\mathrm{crystallization}$$

Answered by MikeH last updated on 14/Oct/22

$$\mathrm{mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=4.5\mathrm{g}$$$$\mathrm{Empirical}\mathrm{Formula}$$$${\mathrm{CuSO}}_4{\mathrm{H}}_2\mathrm{O}$$$$\mathrm{moles}:\frac{4.0}{64+32+4\left(16\right)}\frac{4.5}{18}$$$$=0.0250.25$$$$\mathrm{mole}\mathrm{ratio}:\frac{0.025}{0.025}\frac{0.25}{0.025}$$$$=110$$$$\Rightarrow {\mathrm{CuSO}}_4.10{\mathrm{H}}_2\mathrm{O}$$

Commented by Spillover last updated on 16/Oct/22

$$thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com