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nSO2=0,1(mol); nO2=0,1(mol)
a) PTHH: 2 SO2 + O2 \(⇌\) 2 SO3 (xt: V2O5)
Ta có: 0,1/2 < 0,1/1
=> O2 dư, SO2 hết, tính theo nSO2.
b) nSO3=nSO2=0,1(mol)
=> mSO3=0,1.80=8(g)
PTHH: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,1}{1}\) \(\Rightarrow\) Oxi còn dư, Zn p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=0,05\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=0,05\cdot32=1,6\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)
a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)
b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O2 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
\(m_{CuO}=0,5.80=40\left(g\right)\)
a)\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(m\right)\)
\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
tỉ lệ :2 1 1 1
số mol :0,2 0,1 0,1 0,1
\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
b)\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(m\right)\)
\(PTHH:3Fe+2O_2\underrightarrow{ }Fe_3O_4\)
theo phương trình ta có tỉ lệ\(\dfrac{0,2}{3}>\dfrac{0,1}{2}\)=>Fe dư
\(PTHH:3Fe+2O_2\xrightarrow[]{}Fe_3O_4\)
tỉ lệ :3 2 1
số mol :0,15 0,1 0,05
\(m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)
a)
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
b) $n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$
$n_{O_2} = \dfrac{13,44}{22,4} = 0,6(mol)$
Ta thấy :
$n_{Al} : 4 < n_{O_2} : 3$ nên $O_2$ dư
$n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,4(mol)$
$m_{O_2\ dư} = (0,6 - 0,4).32 = 6,4(gam)$
c) $n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$m_{Al_2O_3} = 0,15.102 = 15,3(gam)$
nKMnO4 = 15.8/158 = 0.1 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.1__________________________0.05
nNa = 1.38/23 = 0.06 (mol)
4Na + O2 -to-> 2Na2O
0.06__0.015
mO2 (dư) = ( 0.025 - 0.015) * 32 = 0.32(g)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
Theo PT: \(n_P=2n_{P_2O_5}=0,1\left(mol\right)\)
\(\Rightarrow m_P=0,1.31=3,1\left(g\right)\)
\(n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
c, Có: \(V_{O_2\left(dư\right)}=2,8.15\%=0,42\left(l\right)\)
\(\Rightarrow V_{O_2}=2,8+0,42=3,22\left(l\right)\)
a, PTHH: S + O2 -> (t°) SO2
b, nS = 6,4/32 = 0,2 (mol)
nO2 = 6,72/22,4 = 0,3 (mol)
LTL: 0,2 < 0,3 => O2 dư
nO2 (pư) = nSO2 = nS = 0,2 (mol)
mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g)
c, mSO2 = 64 . 0,2 = 12,8 (g)
a, \(S+O_2\underrightarrow{t^o}SO_2\)
\(nS=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => oxi dư
\(nO_{2\left(dư\right)}=0,1\left(mol\right)\)
\(mO_{2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
\(nSO_2=nS=0,2\left(mol\right)\)
\(mSO_2=0,2.64=12,8\left(g\right)\)
Câu 8:
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,2}{1}\), ta được O2 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,05\left(mol\right)\\n_{H_2O}=n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,15.22,4=3,36\left(l\right)\)
\(m_{H_2O}=0,1.18=1,8\left(g\right)\)
Bạn tham khảo nhé!
Câu 9:
a, PT: \(2R+O_2\underrightarrow{t^o}2RO\)
Theo ĐLBT KL, có: mR + mO2 = mRO
⇒ mO2 = 4,8 (g)
\(\Rightarrow n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
b, Theo PT: \(n_R=2n_{O_2}=0,3\left(mol\right)\)
\(\Rightarrow M_R=\dfrac{19,2}{0,3}=64\left(g/mol\right)\)
Vậy: M là đồng (Cu).
Câu 10:
Ta có: mBaCl2 = 200.15% = 30 (g)
a, m dd = 200 + 100 = 300 (g)
\(\Rightarrow C\%_{BaCl_2}=\dfrac{30}{300}.100\%=10\%\)
⇒ Nồng độ dung dịch giảm 5%
b, Ta có: \(C\%_{BaCl_2}=\dfrac{30}{150}.100\%=20\%\)
⇒ Nồng độ dung dịch tăng 5%.
Bạn tham khảo nhé!
a) \(PTHH:2SO_2+O_2\xrightarrow[V_2O_5]{450^oC}2SO_3\)
\(n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)\\ n_{O_2}=\dfrac{10}{32}=0,3125\left(mol\right)\)
Lập tỉ lệ: \(\dfrac{n_{SO_2}}{2}< \dfrac{n_{O_2}}{1}\left(\dfrac{0,5}{2}< 0,3125\right)\)
=> SO2 hết O2 dư
Theo pt: \(n_{O_2\left(pư\right)}=\dfrac{n_{SO_2}.2}{3}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
\(n_{O_2\left(dư\right)}=0,3125-0,25=0,0625\left(mol\right)\\ m_{O_2}=0,0625.32=2\left(g\right)\)
c) Theo pt, ta có:\(n_{SO_3}=n_{SO_2}=0,5\left(mol\right)\)
\(m_{SO_3}=0,5.80=40\left(g\right)\)