Question

Answer 1

2.

2A+6HCl->2ACl₃+3H₂

0,2--------------------0,3 mol

n H2= \(\dfrac{6,72}{22,4}\)=0,3 mol

=>\(\dfrac{5,4}{A}=0,2\)

=>M_A=27 g\mol

=>Al là Nhôm (Al)

Answer 2

Gọi số mol Fe, Mg là a, b (mol)

=> 56a + 24b = 3,68 (1)

\(n_{NO_2}=\dfrac{5,376}{22,4}=0,24\left(mol\right)\)

Fe⁰ - 3e --> Fe⁺³

a--->3a

Mg⁰ - 2e --> Mg⁺²

b-->2b

2H⁺ + NO₃^- + 1e --> NO₂ + H₂O

0,48<-------0,24<---0,24

Bảo toàn e: 3a + 2b = 0,24 (2)

(1)(2) => a = 0,04; b = 0,06

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{3,68}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,06.24}{3,68}.100\%=39,13\%\end{matrix}\right.\)

\(n_{HNO_3}=n_{H^+}=0,48\left(mol\right)\)

=> \(m_{HNO_3}=0,48.63=30,24\left(g\right)\)