Question

1)Cần lấy bn gam H2SO4 và bn gam dd H2SO4 19,6% để đc 200g dd H2SO4 40%

2)cần lây bn gam dd NaCl 20% và bn gam dd NaCl 30% để pha chế thành 300g dd NaCl 26%

Answer 1

Bài 1:
\(m_{H_2SO_4.40\%}=200\times40\%=80\left(g\right)\)

\(\Rightarrow m_{H_2O}=200-80=120\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4.19,6\%}=\frac{120}{100\%-19,6\%}=149,25\left(g\right)\)

\(\Rightarrow m_{H_2SO_4.19,6\%}=149,25\times19,6\%=29,253\left(g\right)\)

\(\Rightarrow m_{H_2SO_4}=80-29,253=50,747\left(g\right)\)

Answer 2

Bài 2:

Gọi \(m_{ddNaCl.20\%}=x\left(g\right)\Rightarrow m_{NaCl.20\%}=20\%x=0,2x\left(g\right)\)

\(m_{ddNaCl.30\%}=y\left(g\right)\Rightarrow m_{NaCl.30\%}=30\%y=0,3y\left(g\right)\)

\(m_{NaCl.26\%}=300\times26\%=78\left(g\right)\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=300\\0,2x+0,3y=78\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=120\\y=180\end{matrix}\right.\)

Vậy \(m_{ddNaCl.20\%}=120\left(g\right)\)

\(m_{ddNaCl.30\%}=180\left(g\right)\)