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a)
\(SO_3 + H_2O \to H_2SO_4\)
Theo PTHH : \(n_{H_2SO_4} = n_{SO_2} = \dfrac{8}{80} = 0,1(mol)\)
\(\Rightarrow C\%_{H_2SO_4} = \dfrac{0,1.98}{200}.100\% = 4,9\%\)
b)
\(2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\)
Theo PTHH : \(n_{NaOH} = 2n_{H_2SO_4} = 0,2(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,2.40}{4\%} = 200(gam)\)
\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)
Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,05-------->0,1---------->0,05--------->0,05
a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)
b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)
\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)
\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)
\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
1 2 1 1
0,05 0,1 0,05 0,05 (mol)
\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)
\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)
\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)
Câu 3 :
\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
a) Hiện tượng : Xuất hiện kết tủa trắng
Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)
2 1 1 1
0,2 0,1 0,1 0,1
\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)
b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
\(m_{MgSO4}=0,1.120=12\left(g\right)\)
\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)
c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)
\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)
\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)0/0
Chúc bạn học tốt
Câu 4 :
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,2 0,2
\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)
\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)
Chúc bạn học tốt
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
b, \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,1.135}{8+200}.100\%\approx6,49\%\)
a) MgO + 2HCl → MgCl2 + H2O (1)
b) \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
Theo PT1: \(n_{HCl}=2n_{MgO}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{10,95}{200}\times100\%=5,475\%\)
c) Theo PT1: \(n_{MgCl_2}=n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Sigma m_{dd}=6+200=206\left(g\right)\)
\(\Rightarrow C\%_{ddMgCl_2}=\dfrac{14,25}{206}\times100\%=6,92\%\)
d) HCl + KOH → KCl + H2O (2)
Theo PT2: \(n_{KOH}=n_{HCl}=0,3\left(mol\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)
Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
______0,05_____0,1______0,05 (mol)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{3,65}{7,3\%}=50\left(g\right)\)
Ta có: m dd sau pư = 4 + 50 = 54 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,05.135}{54}.100\%=12,5\%\)
a)mH2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g
nHCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)
PTHH:
NaOH+ HCl→ NaCl+ H2O
1 1 1 1
0,4 0,4 0,4 (mol)
⇒mNaOH=0,4.40=16(g)
Nồng độ % của dd NaOH cần dùng là:
C%NaOH=\(\dfrac{16}{200}\) .100%=8%
b)Ta có:mdd spứ=mdd trc pứ=400g
mNaCl=0,4.58,5=23,4g
Nồng độ % dd muối tạo thành sau pứ là:
C%dd NaCl=\(\dfrac{23,4}{400}\) .100%=5,85%