Question

Cho 0,48g nguyên tố R thuộc nhóm II A tâc dụng với 200g dung dịch HCl 3,65% thu được 0,448(l) chất khí ( đktc)

a, xác định R

b, tính C% các chất trong dung dịch sau phản ứng

Answer 1

\(\text{R + 2HCL ->RCL2 +H2}\)

0,02..............................0,02

Ta có :

\(n_{H2}=0,02\left(mol\right)\)

\(M_R=\frac{0,48}{0,02}=24\)

=> R là Mg

b,\(Mg+2HCl\rightarrow MgCl2+H2\)

\(n_{HCl}=\frac{200.3,65\%}{36,5}=0,2\left(mol\right)\)

\(\Rightarrow n_{HCL_{du}}=0,2-0,04=0,16\left(mol\right)\)

\(mdd_{saupu}=0,48+200-0,02.2=200,44\)

\(C\%dd_{MgCl2}=\frac{0,02.95}{200,44}.100\%=0,95\%\)

\(C\%dd_{HCl_{du}}=\frac{0,16.36,5}{200,44}.100\%=2,91\%\)

Answer 2

https://i.imgur.com/wYw6ydP.jpg

Answer 3

R+2HCl--->RCl2+H2

a) n H2=0,448/22,4=0,02(mol)---> m H2=0,04(g)

Theo pthh

n\(R=n_{H2}=0,02\left(mol\right)\)

m\(_R=\frac{0,48}{0,02}=24\left(Mg\right)\)

b)

n HCl=\(\frac{200.3,65}{100}=7,3\left(g\right)\)

n HCl=\(\frac{7,3}{36,5}=0,2\left(mol\right)\)

mà theo pthh

nHCl=2n H2=0,04(mol)

--->HCl dư

m dd sau pư=200+0,48-0,04=200,44(g)

Theo pthh

n MgCl2=n H2=0,02(mol)

C% MgCl2=\(\frac{0,02.95}{200,44}.100\%=0,95\%\)

n HCl dư=0,2-0,04=0,16(mol)

C%HCl=\(\frac{0,16.36,5}{200,44}.100\%=2,91\%\)

Answer 4

\(\text{R + 2HCL ->RCL2 +H2}\)

a/

\(\text{nR=nH2=0.448/22.4=0.02 mol}\)

\(\Rightarrow\)MR=mR/nR=0.48/0.02=24 (g/mol) {Mg}

b/

\(\text{nMgCL2=0.02 mol}\)

\(\Rightarrow\)mctMgCl2=n*M=0.02*(24+35.5*2)=1.9g

\(\Rightarrow\)mddMgCl2=0.48 +200 -0.448 = 200.032 g

\(\text{C%=1.9/200.032 *100 = 0.95%}\)