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
моль
Составим пропорцию из уравнения:
<span>
</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>
С6Н6 --> C6H5OH
n(C6H6) = m(C6H6)/M(C6H6) = 390/78 = 5моль.
n(C6H6) = n(C6H5OH) = 5моль.
m(C6H5OH) = n(C6H5OH)*M(C6H5OH) = 5*94 = 470г.
выход = mпр(C6H5OH)*mт(C6H5OH) = 410/470 = 0,87 или 87%.
CH3 CH3
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CH3 - C = CH - CH - CH3
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Br
5 атомов углерода в молекуле
Mr(NaOH) = 23 + 16 + 1 = 40
W(Na) = Ar(Na) * n / Mr(NaOH) *100% = 23 / 40 * 100% = 57,5
W(O) = Ar(O) * n / Mr(NaOH) *100% = 16 / 40 *100% = 40%
W(H) = Ar(H) *n / Mr(NaOH) *100% = 1 / 40 *100% = 2,5%
Mr(Al2O3) = 27 *2 + 16 *3 = 102
W(Al) = Ar(Al) * n / Mr(Al2O3) *100% =27 *2 / 102 *100% = 53
W(O) = Ar(O) * n / Mr(Al2O3) * 100% = 16 * 3 / 102 *100% = 47%