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a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+65b=12,1\) (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Bảo toàn electron: \(2n_{Fe}+2n_{Zn}=2n_{H_2}\) \(\Rightarrow2a+2b=0,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{12,1}\cdot100\%\approx46,28\%\\\%m_{Zn}=53,72\%\end{matrix}\right.\)
b)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=n_{Zn}=n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(p.ứ\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow\Sigma n_{H_2SO_4}=0,2\cdot110\%=0,22\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,22-0,2\right)\cdot98=1,96\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=211,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{211,7}\cdot100\%\approx7,18\%\\C\%_{ZnSO_4}=\dfrac{16,1}{211,7}\cdot100\%\approx7,61\%\\C\%_{H_2SO_4}=\dfrac{1,96}{22,4}\cdot100\%\approx0,93\%\end{matrix}\right.\)
\(a) 2NaOH + H_2SO_4 \to Na_2SO_4 + H_2O\\ n_{NaOH} = 2n_{H_2SO_4} = 0,1.1.2 = 0,2(mol)\\ \Rightarrow V_{dd\ NaOH} = \dfrac{0,2}{1} = 0,2(lít)\\ b) Na_2SO_3 + H_2SO_4 \to Na_2SO_4 + SO_2 + H_2O\\ n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)\\ V_{SO_2} = 0,1.22,4 = 2,24(lít)\)
\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow n_{NaOH}=0,4.2=0,8\left(mol\right)\\ m_{NaOH}=0,8.40=32\left(g\right)\\ \Rightarrow m_{ddNaOH}=\dfrac{32.100}{25}=128\left(g\right)\\ \Rightarrow V_{ddNaOH}=\dfrac{m_{ddNaOH}}{D_{ddNaOH}}=\dfrac{128}{1,28}=100\left(ml\right)=0,1\left(l\right)\)
a)
Gọi : \(\left\{{}\begin{matrix}n_{NaCl}=a\left(mol\right)\\n_{KI}=b\left(mol\right)\end{matrix}\right.\)
NaCl + AgNO3 → AgCl + NaNO3
a..............a...............a..............................(mol)
KI + AgNO3→ AgI + KNO3
b.......b..............b..................................(mol)
Ta có :
\(n_{AgNO_3} = a + b = 0,25.2 = 0,5(mol)\)
\(m_{kết\ tủa} = 143,5a + 235b = 103,775\)(gam)
Suy ra : a = 0,15 ; b = 0,35
Vậy :
\(C_{M_{NaCl}} = \dfrac{0,15}{0,4} = 0,375M\\ C_{M_{KI}} = \dfrac{0,35}{0,4} = 0,875M\)
b)
Sau phản ứng, dung dịch gồm : \(\left\{{}\begin{matrix}NaNO_3:0,15\left(mol\right)\\KNO_3:0,35\left(mol\right)\end{matrix}\right.\)
Suy ra :
\(m_{NaNO_3} = 0,15.85 = 12,75(gam)\\ m_{KNO_3} = 0,35.101 = 35,35(gam)\)
Đáp án C
Vì H2S và SO2 đều phản ứng với Br2 sinh ra H2SO4 nên ta chỉ cần bảo toàn lưu huỳnh:
Ta có
\(m_{BaCl_2}=\frac{200.20,8}{100}=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\frac{41,6}{208}=0,2\left(mol\right)\)
\(m_{Na_2SO_4}=\frac{150.28,4}{100}=42,6\left(g\right)\Rightarrow n_{Na_2SO_4}=\frac{42,6}{142}=0,3\left(mol\right)\)
\(BaCl_2+Na_2SO_4\rightarrow BaSO_4\downarrow+2NaCl\)
Thấy \(n_{BaCl_2}< n_{Na_2SO_4}\Rightarrow\) Tính theo \(BaCl_2\)
\(m_{ddsaupu}=200+150=350\left(g\right)\)
Có \(n_{NaCl}=2n_{BaCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaCl}=58,5.0,4=23,4\left(g\right)\Rightarrow C\%\left(NaCl\right)=\frac{23,4}{350}.100\%\approx6,67\%\)
BaCl2 + Na2SO4-----> BaSO4 +2 NaCl
Ta có
m\(_{BaCl2}=\)\(\frac{200.20,8}{100}=41,6\left(g\right)\)
n\(_{BaCl2}=\frac{41,6}{110}=0,38\left(mol\right)\)
m\(_{Na2SO4}=\frac{150.28,4}{100}=42,6\left(g\right)\)
n\(_{Na2SO4}=\frac{42,6}{142}=0,3\left(mol\right)\)
=> BaCl2 dư
Theo pthh
n\(_{BaCl2}=n_{Na2SO4}=0,3\left(Mol\right)\)
n\(_{BaCl2}dư=0,38-0,3=0,08\left(mol\right)\)
mdd= 200+150=350(g)
C%(BaCl2)=\(\frac{0,08.110}{350}.100\%=2,5\%\)
Theo pthh
n\(_{NaCl}=2n_{Na2SO4}=0,6\left(mol\right)\)
C%(NaCl)=\(\frac{0,6.58,5}{350}.100\%=10\%\)
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