ddHCl có C% là bao nhiêu :)?

a) \(n_{H_2SO_4}=\dfrac{5,88}{98}=0,06\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

            0,04<--0,06------->0,02---------->0,06

\(\Rightarrow\left\{{}\begin{matrix}a=m_{Al}=0,04.27=1,08\left(g\right)\\V=V_{H_2}=0,06.22,4=1,344\left(l\right)\end{matrix}\right.\)

b) 

Cách 1: \(m=m_{Al_2\left(SO_4\right)_3}=0,02.342=6,84\left(g\right)\)

Cách 2: \(m_{H_2}=0,06.2=0,12\left(g\right)\)

Áp dụng ĐLBTKL:

\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Rightarrow m=m_{Al_2\left(SO_4\right)_3}=1,08+5,88-0,12=6,84\left(g\right)\)

c) \(n_{O_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{0,06}{1}\Rightarrow\) O₂ dư, H₂ hết

Theo PTHH: \(n_{O_2\left(p\text{ư}\right)}=\dfrac{1}{2}.n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(d\text{ư}\right)}=0,06-0,03=0,03\left(mol\right)\\m_{O_2\left(d\text{ư}\right)}=0,03.32=0,96\left(g\right)\\V_{O_2\left(d\text{ư}\right)}=0,03.22,4=0,672\left(l\right)\end{matrix}\right.\)

Theo PTHH: \(n_{H_2O}=n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,06.18=1,08\left(g\right)\)

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

______0,2-->0,6------->0,2---->0,3

=> m_AlCl3 = 0,2.133,5 = 26,7(g)

b) V = 0,3.22,4 = 6,72(l)

c) Số phân tử HCl = 0,6.6.10²³ = 3,6.10²³

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(\text{mol}\right)\)

Phương trình hóa học phản ứng

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

_{2          :   3                   : 1               : 3} 

0,4        0,6            0,2             0,6

mol       mol           mol             mol

Thể tích khí H₂ sinh ra là 

\(V=n.22,4=0,6.22,4=13,44\left(l\right)\)

\(m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\)

\(m_{H_2SO4}=n.M=0,6.98=58,8\left(g\right)\)

\(m_{H_2}=n.M=0,6.2=1,2\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{58,8.100\%}{9,8\%}=600\)(g)

=> \(m_{\text{dd sau pư}}=m_{ddH_2SO_4}+m_{Al}-m_{H_2}\)

= 600 + 10,8 - 1,2 (g) = 609,6 (g) 

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{m_{Al_2\left(SO_4\right)_3}}{m_{\text{dd sau pư}}}.100\%=\dfrac{68,4}{609,6}.100\%\)=11,22%

\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)

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

\(.........0.6......0.2.......0.3\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)

a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)

a) Pt: $Fe+2HCl\to FeC{}^{}+{}^{}$

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

b, Ta có: \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

Bạn tham khảo nhé!

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,2mol\\n_{H_2}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)

a)

2Al  +   6HCl  →  2AlCl₃  +  3H₂

b) nAl  = 5,4 : 27 = 0,2 mol

Theo tỉ lệ phản ứng => nH₂ = 0,3 mol <=> VH₂ = 0,3.22,4 = 6,72 lít.

c) nAlCl₃ = nAl = 0,2 mol

=> mAlCl₃ = 0,2. 133,5 = 26,7 gam.

d) nHCl cần dùng = 3nAl = 0,6 mol

=> mHCl = 0,6.36,5 = 21,9 gam

<=> mdd HCl cần dùng = \(\dfrac{21,9}{3,65\%}\) = 600 gam