Дано:m(р-ра(NaOH))=20г.,ω(<span>NaOH)=10<span>%, m(AlCl3)-?
x 0,05</span></span>
AlCl3+3NaOH=Al(<span>OH)3+3NaCl
</span><span> 1 3
</span>m(NaOH)=m(р-ра(NaOH))÷100×ω(NaOH)=10%=20÷100×10=2г
![n(NaOH)= \frac{m(NaOH)}{M(NaOH)} = \frac{m(NaOH)}{Mr(Na)+Mr(O)+Mr(H)} = \frac{2}{23+16+1} = \frac{2}{40} =0,05](https://tex.z-dn.net/?f=n%28NaOH%29%3D+%5Cfrac%7Bm%28NaOH%29%7D%7BM%28NaOH%29%7D+%3D+%5Cfrac%7Bm%28NaOH%29%7D%7BMr%28Na%29%2BMr%28O%29%2BMr%28H%29%7D+%3D+%5Cfrac%7B2%7D%7B23%2B16%2B1%7D+%3D+%5Cfrac%7B2%7D%7B40%7D+%3D0%2C05)
моль
Составим пропорцию из уравнения:
<span>
![\frac{x}{1} = \frac{0,05}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%7D%7B1%7D+%3D+%5Cfrac%7B0%2C05%7D%7B3%7D+)
</span>x=0,016 моль <span>AlCl3
</span>m(AlCl3)=M×n=(Mr(Al)+3×Mr(Cl))×n=(27+3×35,5)×0,016=133,5×0,016=2,136г
Ответ:m(AlCl3)=<span>2,136г</span>