Cho 24 gam Fe2O3 vào 250gam dung dịch H2SO4 19,6%. Tính khối lượng của muối thu được sau phản ứng?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nFe2O3=\(\dfrac{24}{160}\)=0,15(mol)
nH2SO4=\(\dfrac{250.19,6\%}{100\%.98}\)=0,5(mol)
PTHH:Fe2O3+3H2SO4→Fe2(SO4)3+3H2O
=>H2SO4 dư
⇒nFe2(SO4)3=nFe2O3=0,15(mol)
⇒mFe2(SO4)3=0,15.400=60(g)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ m_{H_2SO_4}=19,6\%.250=49\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
LTL: \(\dfrac{0,15}{1}< \dfrac{0,5}{3}\rightarrow\) H2SO4 dư
Theo pthh: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\)
\(\rightarrow m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)
Bài 19 :
\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)
Bài 18 :
\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)
nH2 = 6.72 / 22.4 = 0.3 (mol)
Mg + H2SO4 => MgSO4 + H2
0.3.......0.3.............0.3........0.3
mMg = 0.3 * 24 = 7.2 (g)
mH2SO4 = 0.3 * 98 = 29.4 (g)
mddH2SO4 = 29.4 * 100 / 19.6 = 150 (g)
mMgSO4 = 0.3 * 120 = 36 (g)
\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,9(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,9.98}{19,6\%}=450(g)\)
\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.4......................0.4\)
\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)
\(b.\)
\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(0.16..........0.16..............0.16\)
\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)
\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)
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%
Tóm tắt
\(V_{H_2\left(đktc\right)}=8,96l\\ C_{\%H_2SO_4}=19,6\%\\ a)m_{Zn}=?\\ m_{ddH_2SO_4}=?\\ b)C_{\%ZnSO_4}=?\)
\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,4 0,4 0,4 0,4
\(m_{Zn}=0,4.65=26g\\ m_{ddH_2SO_4}=\dfrac{0,4.98}{19,6}\cdot100=200g\\ b)C_{\%ZnSO_4}=\dfrac{0,4.161}{26+200-0,4.2}\cdot100=28,6\%\)
\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15(mol)\\ n_{H_2SO_4}=\dfrac{250.19,6\%}{100\%.98}=0,5(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\)
Vì \(\dfrac{n_{Fe_2O_3}}{1}<\dfrac{n_{H_2SO_4}}{3}\) nên \(H_2SO_4\) dư
\(\Rightarrow n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0,15(mol)\\ \Rightarrow m_{Fe_2(SO_4)_3}=0,15.400=60(g)\)