Câu 378:
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{N_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
BT e, có: \(3n_{Al}=10n_{N_2}+8n_{NH_4NO_3}\)
\(\Rightarrow n_{NH_4NO_3}=0,05\left(mol\right)\)
1. BTNT Al: nAl(NO3)3 = nAl = 0,4 (mol)
⇒ m = mAl(NO3)3 + mNH4NO3 = 0,4.213 + 0,05.80 = 89,2 (g)
→ Đáp án: D
2. nHNO3 = 12nN2 + 10nNH4NO3 = 1,46 (mol)
\(\Rightarrow V_{HNO_3}=\dfrac{1,46}{2}=0,73\left(l\right)=730\left(ml\right)\)
Câu 379:
\(n_{Al}=\dfrac{12,42}{27}=0,46\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{N_2O}+n_{N_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\\44n_{N_2O}+28n_{N_2}=18.2.0,06\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,03\left(mol\right)\\n_{N_2}=0,03\left(mol\right)\end{matrix}\right.\)
BT e, có: \(3n_{Al}=8n_{N_2O}+10n_{N_2}+8n_{NH_4NO_3}\)
\(\Rightarrow n_{NH_4NO_3}=0,105\left(mol\right)\)
1. BTNT Al: nAl(NO3)3 = nAl = 0,46 (mol)
⇒ m = mAl(NO3)3 + mNH4NO3 = 0,46.213 + 0,105.80 = 106,38 (g)
→ Đáp án: C
2. nHNO3 = 10nN2O + 12nN2 + 10nNH4NO3 = 1,71 (mol)
→ Đáp án: A
