a)
$n_K = \dfrac{3,9}{39} = 0,1(mol)$
$2K + 2H_2O \to 2KOH + H_2$
Theo PTHH :
$n_{H_2} = \dfrac{1}{2}n_K = 0,05(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$
b)
$m_{dd\ sau\ pư} = 3,9 + 200 -0,05.2 = 203,8(gam)$
$n_{KOH} = n_K = 0,1(mol)$
$C\%_{KOH} = \dfrac{0,1.56}{203,8}.100\% = 2,75\%$
c)
$2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{KOH} = 0,05(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,05.98}{25\%} = 19,6(gam)$
$V_{dd\ H_2SO_4} = \dfrac{19,6}{1,14}= 17,19(ml)$
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