Ta có: \(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

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

_____0,8_____2,4_____________1,2 (mol)

a, m_HCl = 2,4.36,5 = 87,6 (g)

b, V_H2 = 1,2.24,79 = 29,748 (l)

ko sao dau

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, Ta có: \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,75\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,75.24,79=18,5925\left(l\right)\)

c, \(n_{HCl}=3n_{Al}=1,5\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{1,5}{1}=1,5\left(l\right)=1500\left(ml\right)\)

d, \(n_{AlCl_3}=n_{Al}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,5}{1,5}=\dfrac{1}{3}\left(M\right)\)

C:H:O

40/12 : 6,67/1 : 53,33/16

1:2:1=>GĐGN:ch2o

=>CTPT:(ch2o)n=180->n=6(c6h12o6)

Giúp mình với 😩😩😩

H( Hydrogen)

O( Oxygen)

F(Flourine)

S( Sulfur)

Si( Silicon)

chịu

tui hổng bt

bó tay

Ta có: 392n_Cr2(SO4)3 + 152n_FeSO4 = 67,2 (1)

PT: \(2Cr+3H_2SO_4\rightarrow Cr_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=3n_{Cr_2\left(SO_4\right)_3}+n_{FeSO_4}=0,32.1,5=0,48\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Cr_2\left(SO_4\right)_3}=0,09\left(mol\right)\\n_{FeSO_4}=0,21\left(mol\right)\end{matrix}\right.\)

Theo PT: n_Cr = 2n_Cr2(SO4)3 = 0,18 (mol)

n_Fe = n_FeSO4 = 0,21 (mol)

⇒ n_Cr:n_Fe = 0,18:0,21 = 6:7

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{36,5.20\%}{36,5}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)