Question

cho 2,4 g Mg tác dụng vừa đủ với m gam dd HCl 20% .biết D=1,1g/ml

a) tính thể tính khí thu được

B) tinh M gam dd HCl tham gia phản ứng

c) tính \(c_m\), C% dd sau phản ứng

Answer 1

Ta co pthh

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

Theo de bai ta co

nMg=\(\dfrac{2,4}{24}=0,1\left(mol\right)\)

a,Theo pthh

nH2=nMg=0,1 mol

\(\Rightarrow\) VH2\(_{\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b, Theo pthh

nHCl=2nMg=2.0,1=0,2 mol

\(\Rightarrow mct=mHCl=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow mdd_{HCl}=\dfrac{mct.100\%}{C\%}=\dfrac{7,3.100\%}{20\%}=36,5\left(g\right)\)

c, Theo Pthh ta co

nMgCl2=nMg=0,1 mol

Theo de bai ta co

V\(dd_{HCl}=\dfrac{mdd}{D_{dd}}=\dfrac{36,5}{1,1}\approx33,18ml=0,3318l\)

\(\Rightarrow CM=\dfrac{n}{V}=\dfrac{0,1}{0,3318}\approx0,30M\)

Answer 2

n_Mg=m/M=2,4/24=0,1(mol)

PT:

Mg + 2HCl -> MgCl₂ +H₂

1..........2................1.........1 (mol)

0,1 -> 0,2 -> 0,1 -> 0,1 (mol)

Khí thu được là H₂

V_H2=n.22,4=0,1.22,4=2,24(lít)

b) m_HCl=n.M=0,2.36,5=7,3(gam)

=> \(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{7,3.100}{20}=36,5\left(g\right)\)

c) m_{d d sau phản ứng}=m_Mg + m_HCl-m_H2=2,4+36,5-(0,1.2)=38,7(g)

m_MgCl2=n.M=0,1.95=9,5(g)

\(\Rightarrow C\%_{ddsauphanung}=\dfrac{m_{MgCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{9,5.100}{38,7}\approx24,55\left(\%\right)\)

Ta có: \(V_{ddHCl}=\dfrac{m_{ddHCl}}{D}=\dfrac{36,5}{1,1}\approx33,18\left(ml\right)\)= 0,3318(lít)

=> C_M=\(\dfrac{n}{V}=\dfrac{0,1}{0,3318}\approx0,3\left(M\right)\)

Answer 3

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

\(n_{Mg}=\dfrac{2,4}{24}=0,1mol\)

theo PTHH=> \(n_{H_2}=n_{Mg}=0,1mol\)

a) \(V_{H_2}=0,1.22,4=2,24lit\)

b) theo PTHH=> \(n_{HCl}=2.n_{Mg}=2.0,1=0,2mol\)

=> m_HCl=0,2.36,5=7,3gam

mdd HCl tham gia phản ứng là:\(\dfrac{7,3}{20}.100=36,5gam\)

c) mdd sau phản ứng:2,4+36,5-0,1.2=38,7gam

theo PTHH => \(n_{MgCl_2}=n_{Mg}=0,1mol\)

\(C\%=\dfrac{0,1.95}{38,7}.100\%=24,55\%\)

\(C_M=\dfrac{10.D.C\%}{M}=\dfrac{10.1,1.24,55}{95}=2,84M\)