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Bài 2:
a) [tex]n_{Na}=\dfrac{2,3}{23}=0,1(mol)[/tex]
[tex]2Na+2H_2O\rightarrow 2NaOH+H_2[/tex]
0,1 ----------------------- 0,1 ----------- 0,05 (mol)
Bảo toàn khối lượng:
[tex]m_{dd}(sau pứ)=2,3+197,8-0,05.2=200(g)[/tex]
[tex]\Rightarrow C\%(NaOH)=\dfrac{0,1.40}{200}.100\%=2\%[/tex]
Có [tex]V_{dd}=\dfrac{m_{dd}}{d}=\dfrac{200}{1,08}\approx 185(ml)=0,185(l)[/tex]
[tex]\Rightarrow C_{M}(NaOH)=\dfrac{0,1}{0,185}\approx 0,54(M)[/tex]
b) [tex]m_{dd}=500.1,1=550(g)[/tex]
[tex]\Rightarrow m_{NaOH}=550.20\%=110(g)[/tex]
[tex]\Rightarrow n_{NaOH}=\dfrac{110}{40}=2,75(mol)[/tex]
c) [tex]n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{50}{250}=0,2(mol)[/tex]
[tex]m_{dd}(Sau pứ)=50+450=500(g)[/tex]
[tex]\Rightarrow C\%(CuSO_4)=\dfrac{0,2.160}{500}.100\%=6,4\%[/tex]
Có [tex]V_{dd}=\dfrac{m_{dd}}{d}=\dfrac{500}{1,05}\approx 476(ml)=0,476(l)[/tex]
[tex]\Rightarrow C_{M}(CuSO_4)=\dfrac{0,2}{0,476}\approx 0,42(M)[/tex]
Bài 3:
[tex]m_{ddNa_2CO_3}=V.d(g)[/tex]
[tex]\Rightarrow m_{ddX}=m_{Na_2CO_3.10H_2O}+m_{ddNa_2CO_3}=m+V.d(g)[/tex]
[tex]n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}=\dfrac{m}{286}(mol)[/tex]
[tex]m_{Na_2CO_3}(ddNa_2CO_3)=V.d.C(g)[/tex]
[tex]\Rightarrow \sum m_{Na_2CO_3}=106.\dfrac{m}{286}+V.d.C=\dfrac{53}{143}m+V.d.C(g)[/tex]
[tex]\Rightarrow C\%(X)=\dfrac{\dfrac{53}{143}m+V.d.C}{m+V.d}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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