HOC24
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\(n_{CO_2}=\dfrac{4}{44}=\dfrac{1}{11}mol\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{1}{11}mol\\ m_{CaCO_3}=\dfrac{1}{11}\cdot100=\dfrac{100}{11}g\\ \%CaCO_3=\dfrac{100:11}{10}\cdot100\%=90,91\%\)
\(a.n_{NaOH}=\dfrac{20.20\%}{100\%.40}=0,1mol\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,05mol\\ m_{ddCuSO_4}=\dfrac{0,05.160}{10\%}\cdot100\%=80g\\ b.m_{Cu\left(OH\right)_2}=0,05.98=4,9g\\ c.C_{\%Na_2SO_4}=\dfrac{0,05.142}{20+80-4,9}\cdot100\%=7,46\%\)
\(V_{C_2H_5OH}=\dfrac{36,8.90}{100}=33,12ml\\ m_{C_2H_5OH}=33,12.0,8=26,496g\\ n_{C_2H_5OH}=\dfrac{26,496}{46}=0,576mol\\ 2C_2H_5OH+2K\rightarrow2C_2H_5OK+H_2\\ n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,288mol\\ V_{H_2}=0,228.22,4=6,4512l\)
Cho hỗn hợp qua dung dịch Brom dư, khí etilen tác dụng với Brom, metan không tác dụng. Khí thoát ra là metan
\(2.\\ a.V_{ddNaOH}=\dfrac{300}{1,2}=250ml=0,25l\\ b.n_{NaOH}=0,25.0,4=0,1mol\\ c.m_{NaOH}=0,1.40=4g\\ d.C_{\%NaOH}=\dfrac{4}{300}\cdot100\%=1,33\%\)
\(1.\\ a.m_{ddKOH}=150.1,25=187,5g\\ b.m_{NaOH}=\dfrac{187,5.5,6\%}{100\%}=10,5g\\ c.n_{KOH}=\dfrac{10,5}{56}=0,1875mol\\ d.150ml=0,15l\\ C_{M_{KOH}}=\dfrac{0,1875}{0,15}=1,25M\)
Ta có:
\(\%X=\dfrac{X.3}{3X+95}\cdot100\%=42.07\%\\ \Rightarrow X\approx23\)
Vậy X là Sodium, Na
\(n_{CaCO_3}=\dfrac{50}{100}=0,5mol\\ Ca\left(OH\right)_2+CO_2\xrightarrow[]{}CaCO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,5mol\\ C_2H_5OH+3O_2\xrightarrow[]{}2CO_2+3H_2O\\ n_{C_2H_5OH}=\dfrac{1}{2}n_{CO_2}=0,25mol\\ m_{C_2H_5OH}=0,25.46=11,5g\\ V_{C_2H_5OH}=\dfrac{11,5}{1}=11,5ml\\ S=\dfrac{11,5}{30}\cdot100=38,33^0\)