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\(\begin{cases} m_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}=19,6(g)\\ m_{BaCl_2}=\dfrac{300.20,8\%}{100\%}=62,4(g) \end{cases} \Rightarrow \begin{cases} n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\\ n_{BaCl_2}=\dfrac{62,4}{208}=0,3(mol) \end{cases}\\ a,PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow +2HCl\)
Vì \(\dfrac{n_{H_2SO_4}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư
\(\Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=0,2.233=46,6(g)\)
\(b,n_{HCl}=n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{CT_{HCl}}=0,2.36,5=7,3(g)\\ m_{dd_{HCl}}=100+300-46,6=353,4(g)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{353,4}.100\%\approx 2,07\%\)
a) mM (1)= 200*20/100= 40g
mM (2)= 300*5/100=15g
mM= 40+15=55g
mdd= 200+300=500g
C%= 55/500*100%= 11%
b) Đặt: VH2SO4 (1)= x (l)
VH2SO4 (2)= y (l)
nH2SO4 (1)= 1.5x mol
nH2SO4 (2)= 0.3y mol
nH2SO4= 1.5x + 0.3y= 0.3*0.5=0.15 (mol) (1)
VH2SO4= x + y = 0.3 l (2)
Giải (1) và (2):
x= 0.05
y= 0.25
VH2SO4 (1)= 0.05l
VH2SO4 (2)= 0.25l
a) \(m_{NaCl}=80\times15\%=12\left(g\right)\)
\(m_{ddNaCl}mới=20+80=100\left(g\right)\)
\(\Rightarrow C\%_{NaCl}mới=\frac{12}{100}\times100\%=12\%\)
b) \(m_{NaCl.20\%}=200\times20\%=40\left(g\right)\)
\(m_{NaCl.5\%}=300\times5\%=15\left(g\right)\)
\(\Rightarrow m_{NaCl}mới=40+15=55\left(g\right)\)
\(m_{ddNaCl}mới=200+300=500\left(g\right)\)
\(\Rightarrow C\%_{NaCl}mới=\frac{55}{500}\times100\%=11\%\)
c) \(m_{H_2SO_4.10\%}=100\times10\%=10\left(g\right)\)
\(m_{H_2SO_4.25\%}=150\times25\%=37,5\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}mới=10+37,5=47,5\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}mới=100+150=250\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}mới=\frac{47,5}{250}\times100\%=19\%\)
Áp dụng quy tắc đường chéo ta có:
a) \(D_1=20g\)
\(D_2=80g\)
\(\frac{D_1}{D_2}=\frac{20}{80}=\frac{15-C\%}{C\%}\rightarrow C\%=12\%\)
b) \(D_1=200\left(g\right)\)
\(D_2=300\left(g\right)\)
\(\frac{D_1}{D_2}=\frac{200}{300}=\frac{C\%-5}{20-C\%}\rightarrow C\%=11\%\)
c) \(D_1=100\left(g\right)\)
\(D_2=150\left(g\right)\)
\(\frac{D_1}{D_2}=\frac{100}{150}=\frac{25-C\%}{C\%-10}\rightarrow C\%=19\%\)
Ta có:
n H 2 SO 4 = 0 , 2 x 2 , 5 + 0 , 1 x 1 = 0 , 6 ( mol )
→ C M sau khi trộn = 0,6/0,3 = 2M.
Ta có: n H 2 SO 4 = 0 , 2 x 2 , 5 + 0 , 1 x 1 = 0 , 6 ( mol )
→ C M sau khi trộn = 0 , 6 / 0 , 3 = 2 M .
1) Ta có: \(m_{H_2SO_4}=200\cdot15\%+300\cdot25\%=105\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{105}{200+300}\cdot100\%=21\%\)
2) Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{105}{98}=\dfrac{15}{14}\left(mol\right)\\V_{ddH_2SO_4}=\dfrac{500}{1,25}=400\left(ml\right)\end{matrix}\right.\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{15}{14}}{0,4}\approx2,68\left(M\right)\)
\(V_{\text{dd}}=200+250=450ml=0,45l\\ n_{H_2SO_4}=\left(0,2.1\right)+\left(2.0,25\right)=0,7\left(mol\right)\\ C_M=\dfrac{0,7}{0,45}=1,5M\)
