Câu 2:
Theo ĐLBT KL, có: mR + mCl2 = m muối
⇒ mCl2 = 26,625 (g) ⇒ nCl2 = 0,375 (mol)
PT: \(2R+Cl_2\underrightarrow{t^o}2RCl\)
___0,75_0,375 (mol)
\(\Rightarrow M_R=\dfrac{17,25}{0,75}=23\left(g/mol\right)\)
Vậy: R là Natri (Na).
Câu 3:
Ta có: \(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
a, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 24y = 14,1 (1)
Theo ĐLBT e, có: 3x + 2y = 0,7.2 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,3.27}{14,1}.100\%\approx57,45\%\\\%m_{Mg}\approx42,55\%\end{matrix}\right.\)
b, BTNT H, có: \(n_{HCl}=2n_{H_2}=1,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,4.36,5=51,1\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{51,1}{10\%}=511\left(g\right)\)
Bạn tham khảo nhé!
