\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\a, CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{NaOH}=2.0,05=0,1\left(mol\right)\\ b,C_{MddNaOH}=\dfrac{0,1}{0,1}=1\left(M\right)\\ c,n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\\ m_{muối}=m_{Na_2CO_3}=0,05.106=5,3\left(g\right)\)

\(a.CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ b.n_{CO_2}=\dfrac{1,12}{22,4}=0,05mol\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,05       0,1                  0,05

\(C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1M\\ c.m_{Na_2CO_3}=0,05.106=5,3g\)

a)

$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$
$CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$

c) $BaCl_2 + 2AgNO_3 \to Ba(NO_3)_2 + 2AgCl$

$MgCl_2 + 2AgNO_3 \to 2AgCl + Mg(NO_3)_2$

a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

b. Đổi 100ml = 0,1 lít

Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)

Đáp án D

a) Các chất td được với dd HCl: MgO, KOH, Na, MgSO₄, Fe

Pt: MgO + 2HCl --> MgCl₂ + H₂O

.....KOH + HCl --> KCl + H₂O

....2Na + 2HCl --> 2NaCl + H₂

.....MgSO₄ + 2HCl --> MgCl₂ + H₂SO₄

.....Fe + 2HCl --> FeCl₂ + H₂

b) Các chất td được với CuSO₄: Na, BaCl₂, Fe

Pt: 2Na + 2H₂O --> 2NaOH + H₂

......CuSO₄ + 2NaOH --> Cu(OH)₂ + Na₂SO₄

......BaCl₂ + CuSO₄ --> BaSO₄ + CuCl₂

......Fe + CuSO₄ --> FeSO₄ + Cu

c) Các chất td được với NaOH: HCl, CO₂, MgSO₄

Pt: NaOH + HCl --> NaCl + H₂O

.....\(\left\{{}\begin{matrix}CO_2+NaOH\rightarrow NaHCO_3\\CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\end{matrix}\right.\)

......MgSO₄ + 2NaOH --> Mg(OH)₂ + Na₂SO₄

a)MgO+2HCl--->MgCl2+H2

KOH+HCl--->KCl+H2O

2Na+2HCl--->2NaCl+H2

Fe+2HCl--->FeCl2+H2

b)2Na+2H2O--->2NaOH+H2

2NaOH+CuSO4--->Na2SO4+Cu(OH)2

BaCl2+CuSO4--->BaSO4+CuCl2

Fe+CuSO4--->FeSO4+Cu

c)HCl+NaOH--->NaCl+H2O

CO2+2NaOH--->Na2CO3+H2O

CO2+NaOH--->NaHCO3

MgSO4+2NaOH--->Na2SO4+Mg(OH)2