a) \(PTHH:Mg+2HCL\) → \(MgCl_2+H_2\)

b) Theo định luật bảo toàn khối lượng

⇒ \(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)

c) ⇒ \(4,8+200=m_{MgCl_2}+0,4\)

⇒ \(m_{MgCl_2}=204,4\left(g\right)\)

a, PTHH : \(Mg+2HCl -> MgCl_2+H_2\)

b/ Công thức ĐLBTLKL: \(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)

c/ Áp dụng ĐLBTKL, ta có: \(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)

\(=> m_{MgCl_2}=(9,6+300)-0,8=308,8(g)\)

Vậy khối lượng \(MgCl_2\)  thu được là \(308,8 g \)

a) PTHH: Mg + 2HCl -> MgCl₂ + H₂

b) Theo ĐLBTKL

\(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ =>9,6+300=m_{MgCl_2}+0,8\)

c) Khối lượng MgCl₂ thu dc là

\(m_{MgCl_2}=309,6-0,8=308,8\left(g\right)\)

\(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,4\left(mol\right)\\ m_{MgCl_2}=95.0,4=38\left(g\right)\\ b,V_{H_2\left(đkc\right)}=0,4.24,79=9,916\left(l\right)\\ d,n_{HCl}=0,4.2=0,8\left(mol\right)\\ V_{ddHCl}=\dfrac{0,8}{2}=0,4\left(l\right)\)

a) Mg + 2HCl --> MgCl₂ + H₂

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

            0,2--->0,4-------------->0,2

=> m_HCl = 0,4.36,5 = 14,6 (g)

c) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

d) 

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                       0,2------->0,2

=> m_Cu = 0,2.64 = 12,8 (g)

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(V_{H_2}=0,2.24,79=4,958l\\ c.m_{MgCl_2}=0,2.95=19g\\ d.C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}M\)

a, nMg=0,2(mol)

Mg+2HCl=>MgCl2+H2

b, nH2=nMg=0,2(mol)

=>VH2=4,958(l)

c,nMgCl2=nMg=0,2(mol)

=>mMgCl2=19(g)

d,nHCl=2nMg=0,4(mol)

=>cM(HCl)=0,75(M)

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

0,05 -->0,1----->0,05----->0,05

b

\(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

\(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)

c

\(V_{H_2}=0,05.24,79=1,2395\left(l\right)\)

d

\(V_{HCl}=\dfrac{0,1}{2}=0,05\left(l\right)\)

Bài 14:

Ta có: \(n_{BaCO_3}=\dfrac{39,4}{197}=0,2\left(mol\right)\)

PT: \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)

a, \(n_{CO_2}=n_{BaCO_3}=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.24,79=4,958\left(l\right)\)

b, Sửa đề: tính khối lượng dung dịch HCl → tính nồng độ % dd HCl.

 \(n_{HCl}=2n_{BaCO_3}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\)

c, \(n_{BaCl_2}=n_{BaCO_3}=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 39,4 + 100 - 0,2.44 = 130,6 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2.208}{130,6}.100\%\approx31,85\%\)

Bài 12:

Ta có: \(n_{MgCO_3}=\dfrac{25,2}{84}=0,3\left(mol\right)\)

PT: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

a, Theo PT: \(n_{CO_2}=n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,3.24,79=7,437\left(l\right)\)

b, Ta có: m _{dd sau pư} = 25,2 + 200 - 0,3.44 = 212 (g)

Theo PT: \(n_{MgCl_2}=n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,3.95}{212}.100\%\approx13,44\%\)

Bài 13:

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

1. \(n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\) \(\Rightarrow V_{CO_2}=0,1.24,79=2,479\left(l\right)\)

2. \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}=100\left(g\right)\)

3. Ta có: m _{dd sau pư} = 10 + 100 - 0,1.44 = 105,6 (g)

Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,1.111}{105,6}.100\%\approx10,51\%\)

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