Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
2Al + 6HCl ---> 2AlCl3 + 3H2
Zn + 2HCl ---> ZnCl2 + H2
Theo các pthh trên: \(n_{HCl}=2n_{H_2}=2.0,2=0,4\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}m_{H_2}=0,2.2=0,4\left(g\right)\\m_{HCl}=0,4.36,5=14,6\left(g\right)\end{matrix}\right.\)
Áp dụng ĐLBTKL:
mKim loại + mHCl = mmuối khan + mH2
=> mMuối khan = 12 + 14,6 - 0,4 = 26,2 (g)
pthh fe + 2hcl -> fecl2 + h2
2al2 + 6hcl -> 2l2cl3 + 3h2
zn + 2hcl -> zncl2 + h2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
=> \(n_{H_2SO_4}=0,2\left(mol\right)\)
mmuối = mkim loại + mSO4 = 12 + 0,2.96 = 31,2 (g)
a)
$4Na + O_2 \xrightarrow{t^o} 2Na_2O$
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$2Na + 2HCl \to 2NaCl + H_2$
$Mg + 2HCl \to MgCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
b)
Bảo toàn khối lượng : $m_{O_2} = 4,08 - 2,48 = 1,6(gam)$
$n_{O_2} = \dfrac{1,6}{32} = 0,05(mol)$
Đốt 2,48 gam X cần 0,05 mol $O_2$
Suy ra, đốt 4,96 gam X cần 0,1 mol $O_2$
Mà : \(\dfrac{1}{4}n_{Na}+\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}=n_{O_2}=0,1\)
Theo PTHH :
\(n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Mg}+\dfrac{3}{2}n_{Al}=2\left(\dfrac{1}{4}n_{Na}+\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}\right)=2.0,1=0,2\)$V = 0,2.22,4 = 4,48(lít)$
$n_{HCl} = 2n_{H_2} = 0,4(mol)$
Bảo toàn khối lượng : $m = 4,96 + 0,4.36,5 - 0,2.2 = 19,16(gam)$
\(\left\{{}\begin{matrix}m_{Mg}=\dfrac{40.9}{100}=3,6\left(g\right)\\m_{Al}=9-3,6=5,4\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\end{matrix}\right.\)
PTHH:
Mg + 2HCl ---> MgCl2 + H2
0,15 ------------------------> 0,15
2Al + 6HCl ---> 2AlCl3 + 3H2
0,2 ---------------------------> 0,3
\(\rightarrow V_{H_2}=\left(0,15+0,3\right).22,4=10,08\left(l\right)\)
\(n_{H_2O}=\dfrac{6,48}{18}=0,36\left(mol\right)\)
PTHH: FexOy + yH2 --to--> xFe + yH2O
Theo pthh: \(n_{O\left(oxit\right)}=n_{H_2\left(pư\right)}=n_{H_2O}=0,36\left(mol\right)\)
\(\rightarrow m_{oxit}=15,12+16.0,36=20,88\left(g\right)\)
\(n_{Fe}=\dfrac{15,12}{56}=0,27\left(mol\right)\)
CTHH: FexOy
=> x : y = 0,27 : 0,36 = 3 : 4
=> CTHH: Fe3O4 (oxit sắt từ)
mMg = 40%x9 = 3,6(g) =>nMg=3,6:24 = 0,15 (mol)
=> mAl = 9-3,6 = 5,4(g) => nAl = 5,4:27 = 0,2 (mol)
pthh : 2Al+6HCl -> 2AlCl3+3H2
0,2 0,3
Mg+2HCl -> MgCl2 +H2
0,15 0,15
=> nH2 = 0,15 + 0,3 = 0,45 (mol)
=> VH2 = 0,45.22,4 = 10,08 (L)
mH2 = 0,45 . 2 = 0,9 (mol)
áp dụng BLBTKL ta có :
mH2 + moxit sắt = mFe + mH2O
=> moxit sắt = 20,7 (g)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
=> nHCl = 0,5 (mol)
mmuối = mkim loại + mCl = 10,2 + 0,5.35,5 = 27,95(g)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
=> nHCl = 0,8 (mol)
Theo ĐLBTKL: mkim loại + mHCl = mmuối + mH2
=> mmuối = 15 + 0,8.36,5 - 0,4.2 = 43,4 (g)
Gọi: \(\left\{{}\begin{matrix}n_{H_2}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow x+y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(1\right)\)
Mà: dY/H2 = 6,25
\(\Rightarrow2x+44y=6,25.2.0,4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)=n_{H_2}\\y=0,1\left(mol\right)=n_{CO_2}\end{matrix}\right.\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)
\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m=m_{Al}+m_{Na_2CO_3}=0,2.27+0,1.106=16\left(g\right)\)
Câu 1:
Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)
PT: \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\)
\(CaCO_3+H_2SO_4\rightarrow CaSO_4+H_2O+CO_2\)
\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\)
Theo PT, có: \(n_{H_2O}=n_{CO_2}=n_{H_2SO_4}=0,25\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
Theo ĐLBT KL, có: mhh + mH2SO4 = m muối + mH2O + mCO2
⇒ m muối = mhh + mH2SO4 - mH2O - mCO2
= 25,2 + 0,25.98 - 0,25.18 - 0,25.44
= 34,2 (g)
Bạn tham khảo nhé!
Câu 2:
Ta có: \(n_{H_2SO_4}=0,5\cdot0,75=0,375\left(mol\right)=n_{H_2O}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,375\cdot98=36,75\left(g\right)\\m_{H_2O}=0,375\cdot18=6,75\left(g\right)\end{matrix}\right.\)
Bảo toàn khối lượng: \(m_{oxit}=m_{muối}+m_{H_2O}-m_{H_2SO_4}=28,5\left(g\right)\)
a)
\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\)
b)
Gọi : \(n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a\)
Bảo toàn khối lượng :
\(13,5 + 2a.36,5 = 66,75 + 2.a\\ \Rightarrow a = 0,75\\ \Rightarrow V = 0,75.22,4 = 16,8(lít)\)
a) Mg + 2 HCl -> MgCl2 + H2
2Al + 6 HCl -> 2 AlCl3 + 3 H2
Fe + 2 HCl -> FeCl2 + H2
Zn + 2 HCl -> ZnCl2 + H2
b) mY-mX=mCl
<=> mCl= 66,75-13,5=53,25(g)
=>nCl=53,25/35,5=1,5(mol)
=> nH2= nCl/2= 1,5/2=0,75(mol)
=>V=V(H2,đktc)=0,75.22,4=16,8(l)
\(n_{Na}=\dfrac{4.6}{23}=0,2\left(mol\right)\)
\(n_{Fe}=\dfrac{8.4}{56}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2.7}{27}=0,1\left(mol\right)\)
PTHH : Na + HCl -> NaCl + H2 ( 1 )
0,2 0,2
PTHH : Fe + HCl -> FeCl2 + H2 ( 2 )
0,15 0,15
PTHH : 2Al + 6HCl -> 2AlCl3 + 3H2 ( 3 )
0,1 0,15
\(V_{H_2\left(1\right)}=0,2.22,4=4,48\left(l\right)\)
\(V_{H_2\left(2\right)}=0,15.22,4=3,36\left(l\right)\)
\(V_{H_2\left(3\right)}=0,15.22,4=3,36\left(l\right)\)