\(a.cosB=b.cosA\)

\(\Leftrightarrow a.\dfrac{a^2+c^2-b^2}{2ac}=b.\dfrac{b^2+c^2-a^2}{2bc}\)

\(\Rightarrow a^2+c^2-b^2=b^2+c^2-a^2\)

\(\Leftrightarrow a^2=b^2\)

\(\Rightarrow a=b\)

Hay tam giác ABC cân tại C

Bài thơ "tự tình".

Gọi $n_{H_2} = a(mol) ; n_{CO} = b(mol) ; n_{CO_2} = c(mol)$
$\Rightarrow 2a + 28b + 44c = 3,72(1)$

Mặt khác : 

$n_{hh} = \dfrac{13,44}{22,4} = 0,6(mol)$
$H_2 + CuO \xrightarrow{t^o} Cu + H_2O$
$CO + CuO \xrightarrow{t^o} Cu + CO_2$
$n_{CuO} = n_{H_2} + n_{CO}$

Ta có : $\dfrac{a + b + c}{a + b} = \dfrac{0,6}{0,5}$

$\Rightarrow -0,1a -0,1b + 0,5c = 0(2)$

$C + H_2O \xrightarrow{t^o} CO + H_2$
$C + 2H_2O \xrightarrow{t^o} CO_2 + 2H_2$

Theo PTHH : $n_{H_2} = n_{CO} + 2n_{CO_2}$
$\Rightarrow a = b + 2c(3)$

Từ (1)(2)(3) suy ra : a = 0,14 ; b = 0,06 ; c = 0,04

$n_{H_2O} = n_{H_2} = 0,14(mol)$
$m = 0,14.18 = 2,52(gam)$

\(n_X=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Đặt \(\left\{{}\begin{matrix}n_{H_2}=x\\m_{CO}=y\\n_{CO_2}=z\end{matrix}\right.\) ( mol )

\(\Rightarrow n_X=x+y+z=0,6\left(mol\right)\left(1\right)\)

\(C+H_2O\rightarrow\left(t^o\right)CO+H_2\)

\(C+2H_2O\rightarrow\left(t^o\right)CO_2+2H_2\)

\(Cu+\left\{{}\begin{matrix}CO\\H_2\end{matrix}\right.\rightarrow\left(t^o\right)CuO+\left\{{}\begin{matrix}CO_2\\H_2\end{matrix}\right.\)

\(n_{CuO}=x+y=\dfrac{40}{80}=0,5\left(mol\right)\left(2\right)\)

\(n_{H_2}=n_{CO}+2n_{CO_2}\)

\(\Rightarrow x=y+2z\left(3\right)\)

\(\left(1\right);\left(2\right);\left(3\right)\rightarrow\left\{{}\begin{matrix}x=0,35\\y=0,15\\z=0,1\end{matrix}\right.\)

\(m_X=m_{H_2}+m_{CO}+m_{CO_2}\)

       \(=0,35.2+0,15.28+0,1.44=9,3\left(g\right)\)

Bảo toàn H: \(n_{H_2O}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,35.18=6,3\left(g\right)\)

Ta có: 9,3 gam X `->` 6,3 gam H₂O

           3,72 gam X `->` 2,52 gam H₂O

`=>` \(m=2,52\left(g\right)\)