GS : nA = 1 (mol)
=> nA (pư) = 0.6 (mol)
nA (dư) = 0.4 (mol)
\(C_nH_{2n+2}\underrightarrow{t^0,xt}C_aH_{2a}+C_bH_{2b+2}\left(n=a+b\right)\)
\(M_B=2\cdot18.125=36.25\left(\dfrac{g}{mol}\right)\)
\(m_B=36.25\cdot\left(0.6\cdot2+0.4\right)=58\left(g\right)\)
\(BTKL:m_A=m_B=58\left(g\right)\)
\(M_A=14n+2=58\left(\dfrac{g}{mol}\right)\Rightarrow n=4\)
\(C_4H_{10}\)
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