a)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$
$MgO + 2HCl \to MgCl_2 + H_2O$

$n_{MgCl_2} = n_{MgO} = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)$

b)

$n_{HCl} =2 n_{MgO} = 0,2.2 = 0,4(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,4.36,5}{4\%} = 365(gam)$

a)       MgO +  2HCl→  MgCl₂+  H₂O

(mol)  0,2        0,4         0,2                                                                        

\(n_{MgO}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\)

→\(m_{MgCl_2}=n.M=0,2.95=19\left(g\right)\)

b) Ta có:

\(4\%=\dfrac{m_{HCl_{ }}}{m_{ddHCl}}.100\%< =>4\%=\dfrac{0,4.36,5}{m_{ddHCl}}.100\%\)

=> m_{dd HCl}=\(\dfrac{14,6.100}{4}=365\left(g\right)\)

Vạy khối lượng dung dịch HCl cần dùng cho phản ứng là: 365g

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

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

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

a, \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

b, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)

\(a.\) magnesium+chloric acid→magnesium+hydrogen

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

\(c,\left(b\right)\Leftrightarrow2,4+7,2=9,5+m_{H_2}\\ \Leftrightarrow m_{H_2}=0,1g\)

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

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

\(a,\) Magnesium + Chlorine \(\xrightarrow{t^o}\) Magnesium chloride

\(b,m_{Mg}+m_{Cl_2}=m_{MgCl_2}\\ c,m_{Cl_2}=19-4,8=14,2(g)\)

250ml=0,25l
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0.2.........0.4..........0,2............0,2   (mol)
a)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(m_{MgCl_2}=0,2.95=19\left(g\right)\)
b)
\(C_{M_{HCl}}=\dfrac{0,4}{0,25}=1,6\left(M\right)\)

a/ \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:      0,2       0,4            0,2

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

b/ \(C_{M_{ddHCl}}=\dfrac{0,4}{0,25}=1,6M\)