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\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
Theo PT ta có: \(n_{CaCO_3}=n_{CaCl_2}=n_{CO_2}=0,5\left(mol\right)\)
Theo PT ta có: \(n_{HCl}=\dfrac{0,5.2}{1}=1\left(mol\right)\)
\(\Rightarrow mdd_{HCl}=\dfrac{1.36,5}{20\%}=182,5\left(g\right)\)
\(\Rightarrow mdd_{sau-pư}=m_{CaCO_3}+m_{HCl}-m_{CO_2}\)
\(\Leftrightarrow mdd_{HCl}=50+182,5-22=210,5\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{mct}{mdd}.100\%=\dfrac{0,5.111}{210,5}.100\%\approx26,37\%\)
$n_{NaOH}=0,2.0,5=0,1mol \\PTHH : \\NaOH+HCl\to NaCl+H_2O \\NaOH+HNO_3\to NaNO_3+H_2O \\Gọi\ n_{HCl}=x;n_{HNO_3}=y(x,y>0) \\Ta\ có : \\n_{NaOH}=x+y=0,1mol \\m_{muối}=58,5x+85y=6,38g$
$\text{Ta có hpt :}$
$\left\{\begin{matrix} x+y=0,1 & \\ 58,5x+85y=6,38 & \end{matrix}\right.⇔\left\{\begin{matrix} x=0,08 & \\ y=0,02 & \end{matrix}\right. \\⇒C\%_{HNO_3}=\dfrac{63.0,02}{100}.100\%=1,26\% \\C\%_{HCl}=\dfrac{36,5.0,08}{400}.100\%=0,73\%$
\(m_A=m_{S\left(dư\right)}=0,8\left(g\right)\)
\(\overline{M}_D=9.2=18\)
\(CuCl_2+H_2S\rightarrow CuS+2HCl\)
\(\Rightarrow n_{H2S}=n_{CuS\downarrow}=\frac{9,6}{96}=0,1\left(mol\right)\)
Gọi x là mol H2 trong D
\(\Rightarrow\frac{0,1.34+2x}{0,1+x}=18\)
\(\Leftrightarrow x=0,1\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe\left(dư\right)}=0,1\left(mol\right)\\n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)
\(Fe+S\rightarrow FeS\)
\(\Sigma n_{Fe}=0,1+0,1=0,2\left(mol\right)\)
\(\Rightarrow m_1=0,2.56=11,2\left(g\right)\)
\(\Rightarrow m_2=m_S=0,8+0,1.32=4\left(g\right)\)
Na2O +H2O\(\rightarrow\) 2NaOH
0,5________________1
nNa2O=\(\frac{31}{\text{23.2+16}}\)=0,5mol
500ml=0,5lit
CM NaOH=\(\frac{1}{0,5}\)=2M
2NaOH +H2SO4 \(\rightarrow\)Na2SO4 +2H2O
1__________0,5____0,5
mH2SO4=n.M=0,5.(2+32+16.4)=49g
C%H2SO4=m/mdd. 100
\(\rightarrow\) 20=\(\frac{49}{mdd}\).100
\(\Leftrightarrow\)mddH2SO4=245g
Ta có mddH2SO4=Vdd. D
\(\Leftrightarrow\)245=Vdd.1,14
\(\Leftrightarrow\)VddH2SO4=215ml
215ml=0,215lit
CM Na2SO4=\(\frac{0,5}{0,215}\)=2,3M
PT trung hòa : 2KOH + H2SO4 ----> K2SO4 + 2H2O
---> n H2SO4 = 0.04 (mol) = 1/2 n KOH
---> m H2SO4 = 98 x 0.04 = 3.92 (g)
PT hòa oleum vào nước : H2SO4.nSO3 + nH2O ----> (n+1)H2SO4
áp dụng định luật bảo toàn khối lượng, ta có mH2O = 3.92 - 3.38 = 0.54 (g) -----> n H2O = 0.03 (mol)
lập tỉ số: nH2SO4/nH2O = (n+1)/n = 0.04/0.03= 4/3 ----> n = 3
vậy CT của oleum là H2SO4.3SO3 hay là H2S4O13
1. Gọi CT oleum là \(H_2SO_4.nSO_3\)
\(H_2SO_4.nSO_3+nH_2O\rightarrow\left(n+1\right)H_2SO_4\)
0,015______________________0,015(n+1)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
0,015(n+1)_0,0075(n+1)
\(\Rightarrow n_{NaOH}=0,015\left(n+1\right)=0,01\\ \Rightarrow n=-\frac{2}{3}\)
--------> Sai đề.
2. a) Gọi CT oleum là \(H_2SO_4.nSO_3\)
\(H_2SO_4.nSO_3+nH_2O\rightarrow\left(n+1\right)H_2SO_4\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
0,08_______0,04
\(\Rightarrow n_A=\frac{n_{H_2SO_4}}{n+1}=\frac{0,04}{n+1}\\\Rightarrow M_A=98+80n=\frac{3,38}{\frac{0,04}{n+1}}\\ \Rightarrow n=3\\ \Rightarrow Oleum:H_2SO_4.3SO_3\)
b) Đặt \(n_A=x\left(mol\right)\Rightarrow m_A=338x\left(g\right)\)
\(\Rightarrow m_{dd\text{ }H_2SO_4}=200+338x\left(g\right)\)
\(H_2SO_4.3SO_3+3H_2O\rightarrow4H_2SO_4\)
x________________________4x
\(\Rightarrow m_{H_2SO_4}=98\cdot4x=0,1\left(200+338x\right)\\ \Rightarrow x=0,056\Rightarrow m_A=18,86\left(g\right)\)
Sr bạn nhé Mk tl muộn
RCO3+2HCl\(\rightarrow\)RCl2+CO2+H2O
X2CO3+2HCl\(\rightarrow\)2XCl+CO2+H2O
\(\text{mddHCl=150.1,095=164,25(g)}\)
nHCl=\(\frac{\text{164,25.20%}}{36,5}\)=0,9(mol)
\(\rightarrow\)nhh=0,45(mol)
Mhh=\(\frac{43,3}{0,45}\)=96,2(g)
\(\rightarrow\) R là Mg X là Na
CMHCl=\(\frac{0,9}{0,15}\)=6(M)
m=mhh+mHCl-mCO2-mH2O
\(\text{=43,3+0,9.36,5-0,45.44-0,45.18=48,25(g)}\)
\(CuCl_2+H_2S\rightarrow CuS+2HCl\)
\(m_{dd_{CuCl2}}=3,38.50=169\left(g\right)\)
\(n_{CuCl2}=\frac{169.20\%}{135}=0,25\left(mol\right)\)
\(n_{H2S}=\frac{50.20,4\%}{34}=0,3\left(mol\right)\)
Suy ra H2S dư
\(n_{CuS}=n_{CuCl2}=0,25\left(mol\right)\)
\(n_{HCl}=0,25.2=0,5\left(mol\right)\)
\(m_{dd_{Spu}}=169+50-0,25.96=195\left(g\right)\)
\(C\%_{HCl}=\frac{0,5.36,5}{195}.100\%=9,36\%\)