Question

Answer 1

Câu 381:

\(n_{Al}=\dfrac{12,42}{27}=0,46\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{N_2O}+n_{N_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\\44n_{N_2O}+28n_{N_2}=18.2.0,12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,06\left(mol\right)\\n_{N_2}=0,06\left(mol\right)\end{matrix}\right.\)

BT e, có: \(3n_{Al}=8n_{N_2O}+10n_{N_2}+8n_{NH_4NO_3}\)

\(\Rightarrow n_{NH_4NO_3}=0,0375\left(mol\right)\)

1. BTNT Al: n_Al(NO3)3 = n_Al = 0,46 (mol)

⇒ m = 0,46.213 + 0,0375.80 = 100,98 (g)

→ Đáp án: D

2. n_HNO3 = 10n_N2O + 12n_N2 + 10n_NH4NO3 = 1,695 (mol)

→ Đáp án: A