a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: A + 2HCl ---> ACl2 + H2
0,15<-------------------0,15
=> \(M_A=\dfrac{3,6}{0,15}=24\left(\dfrac{g}{mol}\right)\)
=> A là Mg
b, \(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2
LTL: \(0,15< \dfrac{0,4}{2}\)=> HCl dư
Theo pt: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{H_2}=n_{Mg}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Mg}=2.0,15=0,3\left(mol\right)\end{matrix}\right.\)
mMgCl2 = 0,15.95 = 14,25 (g)
mHCl (dư) = (0,4 - 0,3).36,5 = 3,65 (g)
mH2 = 0,15.2 = 0,3 (g)
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