Hóa học

B

a) 3Fe + 2O₂ --t^o--> Fe₃O₄

Tỉ lệ Fe : O2 = 3:2

b) Áp dụng ĐLBTKL: m_Fe + m_O2 = m_Fe3O4

c) m_Fe + m_O2 = m_Fe3O4

=> m_O2 = 23,2 - 16,8 = 6,4(g)

=> \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)=>V_{O_2}=0,2.22,4=4,48\left(l\right)\)

d) Do Fe : O₂ = 3:2

=> Số phân tử O₂ để đốt cháy hết 0,9.10²³ nguyên tử Fe

= \(\dfrac{2}{3}.0,9.10^{23}=0,6.10^{23}\)

B. 1 và 2 đúng ; 3 sai

\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

B

B

B

a) 2KOH + CO₂ --> K₂CO₃ + H₂O

b)\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 2KOH + CO₂ --> K₂CO₃ + H₂O

______0,1<-------0,05------>0,05_____________(mol)

=> 

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ a,PTHH:2KOH+CO_2\to K_2CO_3+H_2O\\ b,n_{KOH}=2n_{CO_2}=0,1(mol)\\ \Rightarrow C_{M_{KOH}}=\dfrac{0,1}{0,1}=1M\\ c,n_{K_2CO_3}=n_{CO_2}=0,05(mol)\\ \Rightarrow m_{K_2CO_3}=0,05.138=6,9(g)\)

a) 2NaOH + CO₂ --> Na₂CO₃ + H₂O

b) \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 2NaOH + CO₂ --> Na₂CO₃ + H₂O

______0,1<-------0,05------->0,05_____________(mol)

=> \(C_{M\left(NaOH\right)}=\dfrac{0,1}{0,1}=1M\)

c) \(m_{Na_2CO_3}=0,05.106=5,3\left(g\right)\)

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ a,PTHH:2NaOH+CO_2\to Na_2CO_3+H_2O\\ b,n_{NaOH}=2n_{CO_2}=0,1(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1M\\ c,n_{Na_2CO_3}=n_{CO_2}=0,05(mol)\\ \Rightarrow m_{Na_2CO_3}=0,05.106=5,3(g)\)

\(n_{CO_2}=\dfrac{3,7185}{22,4}\left(mol\right);n_{Cl_2}=\dfrac{6,1975}{22,4}\left(mol\right)\)

=> \(m_A=\dfrac{3,7185}{22,4}.44+\dfrac{6,1975}{22,4}.71=26,95\left(g\right)\)

2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O

\(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

B

B

ta có:

\(M+S+4O=160\)

\(M+32+4.16=160\)

\(M+96=160\)

\(M=160-96=64\left(đvC\right)\)

\(\Rightarrow M\) là \(Cu\left(đồng\right)\)

chọn ý B