Câu 384:
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(n_{N_2O}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
1. BT e, có: 3nAl + 2nZn = 8nN2O + 8nNH4NO3
⇒ nNH4NO3 = 0,05 (mol)
BTNT Al: nAl(NO3)3 = nAl = 0,2 (mol)
BTNT Zn: nZn(NO3)2 = nZn = 0,3 (mol)
⇒ m2 = mAl(NO3)3 + mZn(NO3)2 + mNH4NO3 = 103,3 (g)
→ Đáp án: C
2. Có: nHNO3 = 10nN2O + 10nNH4NO3 = 1,5 (mol)
\(\Rightarrow m_{ddHNO_3}=\dfrac{1,5.63}{12,6\%}=750\left(g\right)\)
→ Đáp án: A
Câu 385:
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)=n_{Mg\left(NO_3\right)_2}\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{Zn\left(NO_3\right)_2}\)
\(n_{N_2O}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
1. BT e, có: \(2n_{Mg}+2n_{Zn}=8n_{N_2O}+8n_{NH_4NO_3}\)
\(\Rightarrow n_{NH_4NO_3}=0,05\left(mol\right)\)
\(\Rightarrow m_2=m_{Mg\left(NO_3\right)_2}+m_{Zn\left(NO_3\right)_2}+m_{NH_4NO_3}=71,4\left(g\right)\)
→ Đáp án: C
2. nHNO3 = 10nN2O + 10nNH4NO3 = 1 (mol)
\(\Rightarrow m_1=\dfrac{1.63}{12,6\%}=500\left(g\right)\)
→ Đáp án: C
