Question

Cho Al tác dụng với HCl người ta thu được 6.72 lít khí ở đktc

a, Tính khối lượng Al và HCl đã phản ứng.

b, Tính khối lượng muối.

Answer 1

Answer 2

a) \(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PTHH: \(n_{H_2}:n_{Al}=3:2\)

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

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

Theo PTHH: \(n_{H_2}:n_{HCl}=3:6=1:2\)

\(\Rightarrow n_{HCl}=n_{H_2}.2=0,3.2=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b) Theo PTHH: \(n_{H_2}:n_{AlCl_3}=3:2\)

\(\Rightarrow n_{AlCl_3}=n_{H_2}.\frac{2}{3}=0,3.\frac{2}{3}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

Answer 3

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

a ) Có \(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PTHH \(n_{Al}=\frac{2}{3}n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)

b)

Tương tự ta có

\(n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)

Answer 4

nH2 = 6.72/22.4=0.3 mol

2Al + 6HCl --> 2AlCl3 + 3H2

0.2____0.6_____0.2_____0.3

mAl = 0.2*27=5.4g

mHCl = 0.6*36.5=21.9g

mAlCl3 = 0.2*133.5=26.7g

Answer 5

a) \(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)

PTHH : \(2Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)

Theo PTHH :

\(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}.0,3=0,2mol\) \(\Rightarrow m_{Al}=0,2.27=5,4g\)

\(n_{HCl}=\frac{6}{3}n_{H_2}=2n_{H_2}=2.0,3=0,6mol\) \(\Rightarrow m_{HCl}=0,6.36,5=21,9g\)

b) Theo PTHH : \(n_{Al_2Cl_3}=\frac{2}{3}n_{H_2}=\frac{2}{3}0,3=0,2mol\) \(\Rightarrow m_{Al_2Cl_3}=0,2.133,5=26,7g\)

Answer 6

2Al + 6HCl → 2AlCl₃ + 3H₂

\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

a) theo pt: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,3=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)

b) Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)

Answer 7

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

a) n_{\(H_2\)} = \(\frac{6,72}{22,4}\) = 0,3 mol

Theo PTHH ta có:

n_\(Al\) = \(\frac{0,3.2}{3}\) = 0,2 mol => m_Al = 0,2.27 = 5,4 g

n_HCl = \(\frac{0,3.6}{3}\) = 0,6 mol => m_HCl = 0,6.36,5 = 21,9 g

b)

Theo PTHH, ta có:

n_{\(Al_2Cl_3\)} =\(\frac{0,3.2}{3}\) = 0,2 mol => m_{\(Al_2Cl_3\)} = 0,2.133,5 = 26,7 g

Answer 8

Cảm ơn tất cả mn nha