b,\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂

Mol:     0,2         0,4  

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c,\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂

Mol:      0,2         0,2

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

d,\(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Mol:     0,2           0,4

\(\Rightarrow V_{ddKOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)

\(n_{NaOH}=0.15\cdot1=0.15\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.15...........0.15\)

\(V_{dd_{HCl}}=\dfrac{0.15}{0.5}=0.3\left(l\right)\)

\(C\)

Đổi 150ml = 0,15 lít

Ta có: \(n_{NaOH}=1.0,15=0,15\left(mol\right)\)

PTHH: NaOH + HCl ---> NaCl + H₂O

Theo PT: \(n_{HCl}=n_{NaOH}=0,15\left(mol\right)\)

\(\Rightarrow V_{dd_{HCl}}=\dfrac{0,15}{0,5}=0,3\left(lít\right)\)

2Na + 2 H 2 O → 2NaOH +  H 2

Ba +  2 H 2 O → B a O H 2  +  H 2

Từ 2 phương trình trên ta có nhận xét:

n O H - = 2 n H 2 = 2.33,6/22,4 = 0,15 mol

Mà:  O H - + H + → H 2 O

⇒ n O H -  = n H + = 0,15 mol = nHCl

⇒ V H C l = 0,15/2 = 0,075 lit = 75ml

⇒ Chọn B.

\(n_{HCl}=0,5\cdot0,1=0,05mol\Rightarrow m_{HCl}=0,05\cdot36,5=1,825g\)

\(\Rightarrow m_{ddHCl}=\dfrac{1,825}{36,5\%}\cdot100\%=5g\)

\(\Rightarrow V_{ddHCl}=\dfrac{m_{ddHCl}}{D}=\dfrac{5}{1,18}=4,24ml\)

A

B

a)

$n_{Al_2O_3} = \dfrac{5,1}{102} = 0,05(mol)$
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
$n_{HCl} = 6n_{Al_2O_3} = 0,3(mol)$
$\Rightarrow V = \dfrac{0,3}{4} = 0,075(lít)$

b)

$Al_2O_3 + 2NaOH \to 2NaAlO_2 + 2H_2O$
$n_{NaOH} = 2n_{Al_2O_3} = 0,1(mol)$
$V_{dd\ NaOH} = \dfrac{0,1}{10} = 0,01(lít)$

\(a/KOH+HCl\rightarrow KCl+H_2O\\ n_{HCl}=0,25.1,5=0,375mol\\ n_{KOH}=n_{KCl}=n_{HCl}=0,375mol\\ V_{KOH}=\dfrac{0,375}{2}=0,1875l\\ b/C_{M_{KCl}}=\dfrac{0,375}{0,1875+0,25}=\dfrac{6}{7}M\)

a)

$BaO + H_2O \to Ba(OH)_2$
$n_{Ba(OH)_2} = n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$C_{M_{Ba(OH)_2}} = \dfrac{0,2}{0,2} = 1M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,5} = 0,8(lít)$