a viết pt
b tính % khối lượng mỗi chất trong hỗn hợp
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a)
C2H2 + 2AgNO3 + 2NH3 --> C2Ag2 + 2NH4NO3
C3H6 + Br2 --> C3H6Br2
b)
\(n_{C_2Ag_2}=\dfrac{3,6}{240}=0,015\left(mol\right)\)
=> nC2H2 = 0,015 (mol)
\(m_{tăng}=m_{C_3H_6}=2,1\left(g\right)\)
=> \(n_{C_3H_6}=\dfrac{2,1}{42}=0,05\left(mol\right)\)
\(n_{C_2H_6}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\)
\(\left\{{}\begin{matrix}\%V_{C_2H_6}=\dfrac{0,035}{0,035+0,05+0,015}.100\%=35\%\\\%V_{C_3H_6}=\dfrac{0,05}{0,035+0,05+0,015}.100\%=50\%\\\%V_{C_2H_2}=\dfrac{0,015}{0,035+0,05+0,015}.100\%=15\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{C_2H_6}=\dfrac{0,035.30}{0,035.30+0,05.42+0,015.26}.100\%=29,661\%\\\%m_{C_3H_6}=\dfrac{0,05.42}{0,035.30+0,05.42+0,015.26}.100\%=59,322\%\\\%m_{C_2H_2}=\dfrac{0,015.26}{0,035.30+0,05.42+0,015.26}.100\%=11,017\%\end{matrix}\right.\)
c)
\(n_{CaCO_3}=\dfrac{5}{100}=0,05\left(mol\right)\)
Bảo toàn C: \(n_{CaCO_3}+2.n_{Ca\left(HCO_3\right)_2}=0,25\left(mol\right)\)
=> \(n_{Ca\left(HCO_3\right)_2}=0,1\left(mol\right)\)
Bảo toàn Ca: \(n_{Ca\left(OH\right)_2}=0,15\left(mol\right)\)
=> \(V_{dd.Ca\left(OH\right)_2}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)
Đáp án A
nC2H2=x => nAg2C2=x
nCH3CHO=y => nAg = 2y
+ x+y=11,2/22,4=0,5 (1)
+ 240x+108.2y=112,8 (2)
(1) và (2) => x=0,2; y=0,3
nBr2=2nC2H2+nCH3CHO=2.0,2+0,3=0,7 mol => mBr2=0,7.160=112 gam
nhh = 5.6/22.4 = 0.25(mol)
nAg2C2 = nC2H2 = 24/240 = 0.1 (mol)
nBr2 = 0.3 (mol)
=> nC2H4 = 0.3 - 0.1*2 = 0.1 (mol)
nCH4 = 0.25 - 0.1 - 0.1 = 0.05 (mol)
mCH4 = 0.05*16 = 0.8 (g)
mC2H2 = 0.1*26 = 2.6 (g)
mC2H4 = 0.1*28 = 2.8 (g)
mhh = 0.8 + 2.6 + 2.8 = 6.2 (g)
%CH4 = 0.8/6.2 * 100% = 12.9%
%C2H4 = 2.8/ 6.2 * 100% = 45.16%
%C2H2 = 41.94%
\(n_{C_3H_3Ag}=\dfrac{7,35}{147}=0,05\left(mol\right)\)
=> \(n_{C_3H_4}=0,05\left(mol\right)\)
\(n_{Br_2}=\dfrac{6,4}{160}=0,04\left(mol\right)\)
=> \(n_{C_2H_4}=0,04\left(mol\right)\)
=> \(\left\{{}\begin{matrix}\%C_3H_4=\dfrac{0,05.40}{5,52}.100\%=36,23\%\\\%C_2H_4=\dfrac{0,04.28}{5,52}.100\%=20,29\%\\\%C_2H_6=100\%-36,23\%-20,29\%=43,48\%\end{matrix}\right.\)
=> A