b)Theo PTHH ta có:n\(_{Al_2\left(SO_4\right)_3}\)=\(\dfrac{1}{2}\)n\(_{Al\left(OH\right)_3}\)=\(\dfrac{1}{2}\).0,9487=0,47435(mol)

⇒m\(_{Al_2\left(SO_4\right)_3}\)=0,47435.342=162,2277(g)

a)PTHH: 2Al(OH)₃ + 3H₂SO₄→Al₂(SO₄)₃ + 6H₂O

n\(_{Al\left(OH\right)_3}\)=\(\dfrac{78}{27+\left(16+1\right).3}\)=0,9487(mol)

Theo PTHH ta có: n\(_{H_2SO_4}\)=\(\dfrac{3}{2}\)n\(_{Al\left(OH\right)_3}\)=\(\dfrac{3}{2}\).0,9487=1,42305(mol)

⇒m\(_{H_2SO_4}\)=1,42305.98=139,4589(g)

b)Áp dụng ĐLBTKL, ta có:

m\(_{Al\left(OH\right)_3}\)+ m\(_{H_2SO_4}\)=m\(_{Al_2\left(SO_4\right)_3}\)

⇔m\(_{Al_2\left(SO_4\right)_3}\)=78+139,4589=217,4589(g)

a)Theo PP đường chéo, ta có:

\(\dfrac{2}{3}\)=\(\dfrac{5-C_MC}{C_MC-2}\)⇒C_MC=0,38M

bÁp dụng PP đường chéo, ta có:

\(\dfrac{V_A}{V_B}\)=\(\dfrac{0,5-0,3}{0,3-0,2}\)=2

Vậy phải trộn A và B theo thể tích 2:1

n_HCl = 0,3 . 0,5 = 0,15(mol)

Gọi a là V_HCl(1,5M)

b là V_HCl(0,3M)

Ta có :

\(\dfrac{1,5a+1,3b}{1,3}\)=0,5=> 1,5a + 0,3b = 0,15

=> y = \(\dfrac{0,15-1,5a}{0,3}\)

Mà a + b=0,3(l)

=> a +\(\dfrac{0,15-1,5a}{0,3}\) = 0,3

=>a = 0,05(l) = (50ml)

=> b = 250(ml)

Vậy V_HCl(1,5M)=50ml

V_HCl(0,3M)=250ml

a)PTHH : CH₄ + 2O₂\(\underrightarrow{t^o}\)CO₂ +2H₂O

b)n\(_{CH_4}\)=\(\dfrac{6,4}{16}\)=0,4(mol)

Theo PTHH ta có:n\(_{O_2}\)=2n\(_{CH_4}\)=2.0,4=0,8(mol)⇒V\(_{O_2}\)=0,8.22,4=17,92(l)

c)Theo PTHH ta có :n\(_{H_2O}\)=2n\(_{CH_4}\)=2.0,4=0,8⇒m\(_{H_2O}\)=0,8.18=14,4(g)

3)PTHH:4P+5O₂→2P₂O₅

n_P=\(\dfrac{15,5}{31}\)=0,5(mol)

Theo PTHH ta có:n\(_{O_2}\)=\(\dfrac{5}{4}\)n_P=\(\dfrac{5}{4}\).0,5=0,625(mol)⇒V\(_{O_2}\)=0,625.22,4=14(l)

V\(_{O_2}\)=20%.V_kk⇔V\(_{O_2}\)=\(\dfrac{20.V_{kk}}{100}\)

⇒V_kk=100.\(\dfrac{V_{O_2}}{20}\)=100.\(\dfrac{14}{20}\)=70(l)

2.b)m_{hỗn hợp}=(64+32).0,5=48(g)

2.a)

n\(_{SO_2}\)=\(\dfrac{5,6}{22,4}\)=0,25(mol)

n\(_{O_2}\)=\(\dfrac{8}{32}\)=0,25(mol)

⇒n_{hỗn hợp}=n\(_{SO_2}\)+n\(_{O_2}\)=0,25+0,25=0,5(mol)

2.b) m_{hỗn hợp}= 0,25.(64+32)=48(g)

250000

tick nha 

3)PTHH: CaCO₃\(\underrightarrow{t^o}\)CaO +CO₂

Theo PT:_100_____56(g)

Theo đề:__0,8______0,448(tấn)

m\(_{CaCO_3}\)=80%.1=0,8(tấn)

m_CaO(LT)=\(\dfrac{0,8.56}{100}\)=0,448(tấn)

m_CaO(TT)=\(\dfrac{0,448.70}{100}\)=0.3136(tấn)