2)

$n_{NaOH} = 0,2(mol) ; n_{SO_2} = \dfrac{0,3}(mol)$
Ta có : $n_{NaOH} : n_{SO_2} < 1$

Do đó $SO_2$ dư, muối axit tạo thành

$NaOH + SO_2 \to NaHSO_3$
$n_{NaHSO_3} = n_{NaOH} = 0,2(mol)$
$m_{muối} = 104.0,2 = 20,8(gam)$

mn giúp em vs ạ em cần gấp cảm ơn ạ 

1 I wish I had enough money to go abroad

2 I wish I had come to my best friend's birthday

3 I wish he had invited me to drink coffee

4 I wish I had a sister

5 She wish they hadn't been so noisy lats night

6 He wish the bus hadn't been late yesterday

7 I wish you were given this book

8 He wish his house weren't small

9 She wish she hadn't bought this kind of food

10 If only the weather weren't so terrible today

is a big supermarket next to our school

not met him for 3 months

10-day Tet holiday

a: \(\widehat{B}+\widehat{C}=90^0\)

c: Góc kề bù với C bằng tổng của góc A cộng góc B

Câu 4 : 

\(n_{Fe2O3}=\dfrac{24}{160}=0,15\left(mol\right)\)

Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)

          1               3                    1                 3

       0,15                                 0,15

a) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe2\left(SO4\right)3}=0,15.400=60\left(g\right)\)

b) \(C_{M_{Fe2\left(SO4\right)3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

 Chúc bạn học tốt

a,\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:     0,15         0,45                0,15                   

\(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)

b,\(C_{M_{ddFe_2\left(SO_4\right)_3}}=\dfrac{0,15}{0,5}=0,3\left(mol\right)\)

Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O

0,15--------------------------------0,15 mol

n Fe2O3=24\160=0,15 mol

=>m Fe2(SO4)3=0,15.400=60g

mà thể tích khôi đổi 

=>CM Fe2(SO4)3=0,15\0,5=0,3 M

1, The homework is too difficult for him to do

2, The homework isn't easy enough for him to do

3, The homework is so difficult that he can't do it

4, It is such difficult homework that he can't do it