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a. \(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\)
b. Áp dụng định luật bảo toàn khối lượng, ta có:
\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Leftrightarrow m_{Al_2\left(SO_4\right)_3}=m_{Al}+m_{H_2SO_4}-m_{H_2}=5,4+29,4-0,6=34,2\left(g\right)\)
Canxi oxit: CaO : 56đvc
Caxi nitrat : Ca(NO3)2:164đvc
Bari hidroxit:Ba(OH)2:171đvc
Bari sunfat:BaSO4 :233đvc
Lưu huỳnh đioxit: SO2 :64đvc
Kali sunfit: K2SO3:158đvc
đồng hidroxit: Cu(OH)2 : 98đvc
Axxit clohidric: HCl :36,5đvc
Kaliclorua:KCl:74,5đvc
axxit sunfuric:H2SO4 :98đvc
Lưu huỳnh trioxit :SO3 :80đvc
Sắt (III) Clorua: FeCl3 :162,5đvc
Bari nitrat:Ba(NO3)2:261đvc
Đồng (II) oxit: CuO :80đvc
Cabonat:CO3 :60đvc
Nhôm sunfat:Al2(SO4)3:342đvc
Natriphotphat: Na3PO4:164đvc
Magie clo rua: MgCl2 :95đvc
Mangan ddioxxit:MnO2:87đvc
Điphotphopentaoxit:P2O5:142đvc
a+b
- CaO = 56 đvC
Ca(NO3)2 = 164 đvC
Ba(OH)2 : 171 đvC
BaSO4 = 233 đvC
SO2 = 64đvC
K2SO3 = 158đvC
Cu(OH)2=98đvC
HCl= 36,5 đvC
KCl = 74,5 đvC
H2SO4 = 98đvC
Tương tự làm tiếp đi nhé
a) \(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) Theo định luật bảo toàn khối lượng, ta có:
\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\\ \Rightarrow27+m_{H_2SO_4}=171+3\\ \Rightarrow m_{H_2SO_4}=171+3-27=147\left(g\right)\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2--->0,3-------->0,1----------->0,3
b) `V_{H_2} = 0,3.22,4 = 6,72 (l)`
c) `m_{H_2SO_4} = 0,3.98 = 29,4 9g)`
d) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Xét tỉ lệ: 0,2 < 0,3 => H2 dư
\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
nAl = 0,45 mol
nHCl = 2,15
nAl : nHCl = \(\dfrac{0,45}{2}:\dfrac{2,15}{6}=0,225:0,35\)
Do 0,225 < 0,35 nên Al hết, HCl dư
THeo pt: nAlCl3 = nAl = 0,45mol
=> mAlCl3 = 0,45.133,5=60,075g => a = 60,075
Theo pt: nH2 = \(\dfrac{3}{2}nAl=0,675mol\)
=> V = 0,675.22,4 = 15,12l => b = 15,12l
\(n_{H_2}=\dfrac{6}{2}=3\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2..............3.....................1.........3\)
\(m_{Al}=2\cdot27=54\left(g\right)\)
\(m_{H_2SO_4}=3\cdot98=294\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=1\cdot342=342\left(g\right)\)
a) \(2Al+3H_2SO_4\text{ (loãng)}\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(2Al+6H_2SO_4\text{ (đặc)}\xrightarrow[]{t^\circ}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
b) \(3Ca\left(NO_3\right)_2+2Na_3PO_4\rightarrow Ca_3\left(PO_4\right)_2\downarrow+6NaNO_3\)
\(a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ b.3Ca\left(NO_3\right)_2+2Na_3PO_4\rightarrow Ca_3\left(PO_4\right)_2+2NaNO_3\)