Question

pro hóa ơi, giải hộ tớ với :)). Câu 17 nhé

Answer 1

Câu 17:

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

b+c) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)=n_{MgCl_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ddHCl}=\dfrac{0,3\cdot36,5}{14,6\%}=75\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\\m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\end{matrix}\right.\)

d) Ta có: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=78,3\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{14,25}{78,3}\cdot100\%\approx18,2\%\)

Answer 2

a) $Mg + 2HCl \to MgCl_2  + H_2$
b) n H2 = n Mg = 3,6/24  = 0,15(mol)

V H2 = 0,15.22,4 = 3,36 lít

c) n HCl = 2n Mg = 0,3(mol)

=> m dd HCl = 0,3.36,5/14,6% = 75(gam)

d)

n MgCl2 = n Mg = 0,15(mol)

Sau phản ứng : 

m dd = m Mg + mdd HCl - m H2 = 3,6 + 75 - 0,15.2 = 78,3(gam)

C% MgCl2 = 0,15.95/78,3   .100% = 18,2%

Answer 3

\(n_{Mg}=\dfrac{3.6}{24}=0.15\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.15.....0.3............0.15........0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{dd_{HCl}}=\dfrac{0.3\cdot36.5\cdot100}{14.6}=75\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=3.6+75-0.15\cdot2=78.3\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.15\cdot95}{78.3}\cdot100\%=18.2\%\)