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Lê nhật anh hay anh lê nhật?

a) \(n_{HNO_3}=\dfrac{37,8\left(40,3\cdot1,24\right)}{100\cdot63}=0,3\left(mol\right)\)

KOH + HNO₃ \(\rightarrow\)KNO₃ + H₂O

0,3-----0,3----------0,3

\(m_{ddKOH}=\dfrac{56\cdot0,3\cdot100}{33,6}=50\left(g\right)\)

\(m_{ddHNO_3}=40,3\cdot1,24=50\left(g\right)\)

\(m_{KNO_3}=0,3\cdot101=30,3\left(g\right)\)

Dựa vào đề bài ta có phương trình :

\(\dfrac{30,3-m}{\left(50+50\right)-m}=\dfrac{11,6}{100}\Rightarrow m=21,15\left(g\right)\)

b) Dung dịch B là dung dịch bão hòa ở 0^oC

a) PTHH : 2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow n_{H_2}=n_{Al}\cdot\dfrac{3}{2}=0,2\cdot\dfrac{3}{2}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)

b) \(n_{HCl}=n_{Al}\cdot\dfrac{6}{2}=0,2\cdot\dfrac{6}{2}=0,6\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0,6}{1,5}=0,4\left(l\right)\)