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\(n_{H_2SO_4}=1\cdot0,15=0,15\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{Fe}=0,15\cdot56=8,4\left(g\right)\left(B\right)\)
\(m_{ct}=\dfrac{9,8.150}{100}=14,7\left(g\right)\)
\(n_{H2SO4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
\(n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\)
Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O|\)
1 1 1 1
0,25 0,15 0,15
a) Lap ti so so sanh : \(\dfrac{0,25}{1}>\dfrac{0,15}{1}\)
⇒ MgO du , H2SO4 phan ung het
⇒ Tinh toan dua vao so mol cua H2SO4
\(n_{MgSO4}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{MgSO4}=0,15.120=18\left(g\right)\)
b) \(m_{ddspu}=150+10=160\left(g\right)\)
\(C_{MgSO4}=\dfrac{18.100}{160}=11,25\)0/0
Chuc ban hoc tot
Câu 3:
Gọi x, y lần lượt là số mol của MgO và Al2O3
Ta có: \(n_{H_2SO_4}=0,2.250:1000=0,05\left(mol\right)\)
a. PTHH:
MgO + H2SO4 ---> MgSO4 + H2O (1)
Al2O3 + 3H2SO4 ---> Al2(SO4)3 + 3H2O (2)
b. Theo PT(1): \(n_{H_2SO_4}=n_{MgO}=x\left(mol\right)\)
Theo PT(2): \(n_{H_2SO_4}=3.n_{Al_2O_3}=3y\left(mol\right)\)
=> x + 3y = 0,05 (1)
Theo đề, ta có: 40x + 102y = 1,82 (2)
Từ (1) và (2), ta có HPT:
\(\left\{{}\begin{matrix}x+3y=0,05\\40x+102y=1,82\end{matrix}\right.\)
=> x = 0,02, y = 0,01
=> \(m_{MgO}=0,02.40=0,8\left(mol\right)\)
=> \(\%_{m_{MgO}}=\dfrac{0,8}{1,82}.100\%=43,96\%\)
\(\%_{m_{Al_2O_3}}=100\%-43,96\%=56,04\%\)
Câu 4:
Ta có: \(m_{H_2SO_4}=\dfrac{19,6\%.100\%}{100}=19,6\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Ta lại có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a. PTHH: CuO + H2SO4 ---> CuSO4 + H2O
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\)
Vậy H2SO4 dư.
Theo PT: \(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)
=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)
Ta có: \(m_{dd_{CuSO_4}}=8+100=108\left(g\right)\)
=> \(C_{\%_{CuSO_4}}=\dfrac{16}{108}.100\%=14,81\%\)
Câu 5: Thiếu đề
\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.4......................0.4\)
\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)
\(b.\)
\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(0.16..........0.16..............0.16\)
\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)
\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)
a)
$SO_3 + H_2O \to H_2SO_4$
n SO3 = n H2SO4 = 200.19,6%/98 = 0,4(mol)
=> m = 0,4.80 = 32(gam)
b)
$MgO + H_2SO_4 \to MgSO_4 + H_2O$
n MgSO4 = n MgO = n H2SO4 = 80.19,6%/98 = 0,16(mol)
=> m MgO = 0,16.40 = 6,4(gam)
Sau pư, m dd = 6,4 + 80 = 86,4(gam)
=> C% MgSO4 = 0,16.120/86,4 .100% = 22,22%
\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\\
m_{H_2SO_{\text{ 4}}}=\dfrac{100.9,8}{100}=9,8g\\
n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\\
pthh:Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,1 0,1
\(m_{\text{dd}}=10,2+100-\left(0,3.18\right)=104,8g\\
C\%=\dfrac{0,1.342}{104,8}.100\%=32,633\%\)
\(a,n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\\ m_{H_2SO_4}=9,8\%.100=9,8\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
PTHH: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Ban đầu: 0,1 0,1
Phản ứng: \(\dfrac{1}{30}\) 0,1
Sau pư: \(\dfrac{1}{15}\) 0 \(\dfrac{1}{30}\) 0,1
b, \(\rightarrow m_{dd}=\dfrac{1}{30}.102+100=103,4\left(g\right)\)
\(\rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{1}{30}.342}{101,6}.100\%=11,22\%\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ n_{H_2SO_4}=n_{MgO}=0,1\left(mol\right)\\ m_{H_2SO_4}=0,1.98=9,8\left(g\right)\\ m_{ddH_2SO_4}=\dfrac{9,8.100}{9,8}=100\left(g\right)\)
Chọn C.
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ n_{H_2SO_4}=n_{MgO}=0,1\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,1.98.100}{9,8}=100\left(g\right)\\ \Rightarrow ChọnC\)