Bài 1:
a, Phần này đề có thiếu gì không bạn nhỉ?
b, Ta có: mNaOH = 0,5.40 = 20 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{20}{20+500}.100\%\approx3,85\%\)
Bài 2:
Ta có: \(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
___2/15____0,4_____2/15____0,2 (mol)
a, VH2 = 0,2.22,4 = 4,48 (l)
b, mAlCl3 = 2/15.133,5 = 17,8 (g)
c, mAl = 2/15.27 = 3,6 (g)
Bạn tham khảo nhé!
Bài 1:
a, CM KCl = \(\dfrac{n_{KCl}}{V_{dd}}=\dfrac{\text{n}_{KCl}}{2}\)
b, mNaOH = n. M = 0,5 . 40 = 20 (g)
mdd = 20 + 500 = 520 (g)
=> C% = \(\dfrac{20}{520}.100\%=3,85\%\)
1 a) \(CM_{KCl}=\dfrac{n_{KCl}}{2}\left(M\right)\)
b) \(C\%_{NaOH}=\dfrac{0,5.40}{0,5.40+500}.100=3,85\%\)
2.a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H_2}=n_{HCl}=0,2.2=0,4\left(mol\right)\)
=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)
b)\(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=\dfrac{2}{15}\left(mol\right)\)
=> \(m_{AlCl_3}=\dfrac{2}{15}.133,5=17,8\left(g\right)\)
c) \(n_{Al}=\dfrac{1}{3}n_{HCl}=\dfrac{2}{15}\left(mol\right)\)
\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)
