Question

đốt cháy hoàn toàn 33.4 gam hỗn hợp X gồm Al,Fe,Cu ngoài không khí thu được 41,4 gam hỗn hợp Y gồm 3 oxit . cho Y tác dụng hoàn toàn với dung dịch H2SO4 20% biết D=1,14 g/ml , thể tích tối thiểu của dung dịch H2SO4 20% để hoà tan hết hỗn hợp Y là bao nhiu ???

Answer 1

Al,Fe,Cu→+O2" id="MathJax-Element-2-Frame" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" tabindex="0">→+O2→+H2SO4" id="MathJax-Element-3-Frame" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" tabindex="0">→+H2SO4H2O" id="MathJax-Element-4-Frame" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" tabindex="0">H2O
nH2SO4=nH2O=nO41,4−33,416=0,5" id="MathJax-Element-6-Frame" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" tabindex="0">41,4−33,416=0,5 mol
V=md=0,5.98.1001,14.20=215" id="MathJax-Element-7-Frame" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" tabindex="0">V=md=0,5.98.1001,14.20=215 ml

Answer 2

PTHH :

4Fe + 3O2 = 2Fe2O3 (1)

3Fe + 2O2 = Fe3O4 (2)

2Cu + O2 = 2CuO (3)

Fe2O3 + 3H2SO4 = Fe2(SO4)3 + 3H3O (4)

Fe3O4 + 4H2SO4 = FeSO4 + Fe2(SO4)3 + 4H2O (5)

CuO + H2SO4 = CuSO4 + H2O

Áp dụng ĐLBTKL có :

Khối lượng oxi tham gia phản ứng :

\(m_{O_2}=m_{hhY}-m_{hhX}\)

\(\Rightarrow m_O=8\left(g\right)->n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)->n_O=0,05\left(mol\right)\)\(=n_{H_2O}\)\(\Rightarrow\)\(n_{H_2O}=n_{H_2SO_4}=0,5\left(mol\right)\)( theo 4 ,5 )

Mà ta có :

\(V_{dd}=\dfrac{m_{dd}}{D}=\dfrac{0,5.98.100}{1,14.20}=215\left(ml\right)\)

Vậy .....................................