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Câu 1:
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)
c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)
d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)
Câu 2:
a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)
d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)
\(n_{BaCO_3}=\dfrac{19.7}{197}=0.1\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0.15\cdot1=0.15\left(mol\right)\)
\(n_{MgCO_3}=a\left(mol\right),n_{CaCO_3}=b\left(mol\right)\)
\(\Rightarrow m_A=84a+100b=18.4\left(g\right)\left(1\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(n_{CO_2}=a+b\left(mol\right)\)
TH1 : Không tạo muối axit , Ba(OH)2 dư
\(\Rightarrow n_{CO_2}=n_{BaCO_3}=0.1\left(mol\right)\)
\(\Rightarrow a+b=0.1\left(2\right)\)
\(\left(1\right),\left(2\right):a=-0.525,b=0.625\left(L\right)\)
TH2 : Phản ứng tạo hai muối vừa đủ
\(n_{CO_2}=0.1+\left(0.15-0.1\right)\cdot2=0.2\left(mol\right)\)
\(\Rightarrow a+b=0.1\left(3\right)\)
\(\left(1\right),\left(3\right):a=b=0.1\)
\(\%MgCO_3=\dfrac{8.4}{18.4}\cdot100\%=45.65\%\)
\(\%CaCO_3=54.35\%\)
1.
Al2O3 + 2NaOH -> 2NaAlO2 + H2O (1)
nNaAlO2=0,225(mol)
Từ 1:
nNaOH=nNaAlO2=0,225(mol)
nal2O3=\(\dfrac{1}{2}\)nNaAlO2=0,1125(mol)
V dd NaOH=0,225:5=0,045(lít)
mAl2O3=0,1125.102=11,475(g)
mquặng=11,475.110%=12,6225(g)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)
\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)
2.
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
0,2 0,2 0,1
\(m_{KOH}=0,2.56=11,2\left(g\right)\)
\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)
\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_____0,2_____0,6____________0,3 (mol)
a, \(m_{ddHCl}=\dfrac{0,6.36,5}{15\%}=146\left(g\right)\)
b, m dd A = 5,4 + 146 - 0,3.2 = 150,8 (g)
c, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)
d, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
___0,3____0,15 (mol)
\(\Rightarrow V_{O_2}=0,15.24,79=3,7185\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=18,5925\left(l\right)\)
