Câu 1:

\(m_{CuSO_4}=600\times8\%=48\left(g\right)\)

\(\Rightarrow n_{CuSO_4}=\frac{48}{160}=0,3\left(mol\right)\)

Ta có: \(n_{CuSO_4.5H_2O}=n_{CuSO_4}=0,3\left(mol\right)\)

\(\Rightarrow m_{CuSO_4.5H_2O}=0,3\times250=75\left(g\right)\)

Câu 2:

\(m_{CuSO_4.8\%}=500\times8\%=40\left(g\right)\)

\(\Rightarrow m_{H_2O}=500-40=460\left(g\right)\)

\(\Rightarrow m_{ddCuSO_4.4\%}=\frac{460}{100\%-4\%}=479,17\left(g\right)\)

\(\Rightarrow m_{CuSO_4.4\%}=479,17-460=19,17\left(g\right)\)

\(\Rightarrow m_{CuSO_4}\left(trongtt\right)=40-19,17=20,83\left(g\right)\)

\(\Rightarrow n_{CuSO_4}\left(trongtt\right)=\frac{20,83}{160}=0,13\left(mol\right)\)

Ta có: \(n_{CuSO_4.5H_2O}=n_{CuSO_4}\left(trongtt\right)=0,13\left(mol\right)\)

\(\Rightarrow m_{CuSO_4.5H_2O}=0,13\times250=32,5\left(g\right)\)

Bài 3:

\(n_{CuSO_4}=0,2\times0,2=0,04\left(mol\right)\)

Ta có: \(n_{CuSO_4.pH_2O}=0,04\left(mol\right)\)

\(\Rightarrow M_{CuSO_4.pH_2O}=\frac{10}{0,04}=250\left(g\right)\)

\(\Leftrightarrow160+18p=250\)

\(\Leftrightarrow18p=90\)

\(\Leftrightarrow p=5\)

Bài 2:

\(n_{BaCl_2.xH_2O}=\frac{24,4}{208+18x}\left(mol\right)\)

Ta có: \(n_{BaCl_2}=n_{BaCl_2.xH_2O}=\frac{24,4}{208+18x}\left(mol\right)\)

\(\Rightarrow m_{BaCl_2}=\frac{24,4}{208+18x}\times208=\frac{5075,2}{208+18x}\left(g\right)\)

ta có: \(m_{ddBaCl_2}mới=24,4+175,6=200\left(g\right)\)

\(C\%_{BaCl_2}mới=\frac{5075,2}{208+18x}\div200\times100\%=10,4\%\)

\(\Leftrightarrow\frac{5075,2}{41600+3600x}=0,104\)

\(\Leftrightarrow5075,2=4326,4+374,4x\)

\(\Leftrightarrow748,8=374,4x\)

\(\Leftrightarrow x=2\)

Gọi số \(\left(g\right)\) tinh thể \(CuSO_4\cdot5H_2O\) cần pha chế là \(x\left(g\right)\left(0< x< 750\right)\)

Số \(\left(g\right)\) dung dịch \(CuSO_4\text{ }4\%\) cần pha chế là \(y\left(g\right)\left(0< y< 750\right)\)

\(n_{CuSO_4\cdot5H_2O}=\dfrac{m}{M}=\dfrac{x}{250}=0,004x\left(mol\right)\\ \Rightarrow m_{CuSO_4\text{ trong }CuSO_4.5H_2O}=n\cdot M=0,004x\cdot160=0,64x\left(g\right)\)

\(m_{CuSO_4\text{ trong }d^28\%}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{y\cdot4}{100}=0,04y\left(g\right)\)

\(m_{CuSO_4\text{ trong }750\left(g\right)d^28\%}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{750\cdot8}{100}=60\left(g\right)\)

Ta có hệ pt: \(\left\{{}\begin{matrix}x+y=750\\0,64x+0,04y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=50\\y=700\end{matrix}\right.\)

\(\Rightarrow m_{CuSO_4.5H_2O}=50\left(g\right)\\ m_{d^2CuSO_44\%}=700\left(g\right)\)

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khối lượng dung dịch trên =414,594.1,206=500 g

n CuSO4.5H2O= 140,625:250=0,5625 mol

n CuSO4=0,5625 mol

m CuSO4=0,5625.160=90 g

C% CuSO4=90:500=18%

CM CuSO4=0,5625:0,414594=1,37 mol

Tham Khảo

mCuSO4.5H2O = x (gam)

—> mCuSO4 = 160x/250

mdd CuSO4 = y gam

—> mCuSO4 = 0,04y

Khi pha trộn:

x + y = 500

160x/250 + 0,04y = 500*8%

—> x = 33.33 và y = 466.67

Câu 1:

\(m_{NaCl}=500\times10\%=50\left(g\right)\)

Câu 2:

\(m_{CuSO_4}=500\times8\%=40\left(g\right)\)

\(\Rightarrow n_{CuSO_4}=\frac{40}{160}=0,25\left(mol\right)\)

Ta có: \(n_{CuSO_4.5H_2O}=n_{CuSO_4}=0,25\left(mol\right)\)

\(\Rightarrow m_{CuSO_4.5H_2O}=0,25\times250=62,5\left(g\right)\)

\(\Rightarrow m_{H_2O}=500-62,5=437,5\left(g\right)\)

m_{CuSO4 20%} = 54 (g)

m_H2O = 216 (g)

Trong 1 gam tinh thể có:

0.64 (g) CuSO4; 0.36 (g) H2O

Trong 1 gam dd CuSO4 10% có 0.1 (g) CuSO4

Gọi x,y lần lượt là số (g) dd CuSO4 10% và tinh thể cần dùng

Ta có: x + y = 270

0.1x + 0.64y = 54

=> x = 220; y = 50.

Vậy cần dùng 220 g dd CuSO4 10% và 50 g tinh thể để điều chế 270 g dd CuSO4 20%

2a.