Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
PTHH : S + O2 - to---> SO2
0,1 0,1 0,1 ( mol )
b) \(m_S=0,1.32=3,2\left(g\right)\)
\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
a) Có chất mới sinh ra
b) Theo ĐLBTKL: mS + mO2 = mSO2
=> mO2 = 6,4 - 3,2 = 3,2 (g)
c) Xét \(d_{O_2/kk}=\dfrac{32}{29}=1,1\)
=> Khí O2 nặng hơn không khí 1,1 lần
PTHH : \(S+O_2\xrightarrow[]{t^o}SO_2\)
\(BTKl:\) \(m_S+m_{O2}=m_{SO2}\)
\(\Rightarrow m_{O2}=m_{SO2}-m_S=6,4-3,2=3,2\left(g\right)\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
c, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)
\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Theo PT: \(n_{SO_2\left(LT\right)}=n_S=0,1\left(mol\right)\)
\(\Rightarrow m_{SO_2\left(LT\right)}=0,1.64=6,4\left(g\right)\)
Mà: H = 80%
\(\Rightarrow m_{SO_2\left(TT\right)}=6,4.80\%=5,12\left(g\right)\)
\(m_C=1000\cdot90\%=900\left(kg\right)\)
\(n_C=\dfrac{900}{12}=75\left(kmol\right)\)
\(C+O_2\underrightarrow{^{^{t^o}}}CO_2\)
\(75............75\)
\(m_{CO_2}=75\cdot44=3300\left(kg\right)\)
\(m_{CO_2\left(tt\right)}=3300\cdot80\%=2640\left(kg\right)=2.64\left(\text{tấn}\right)\)
$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$
$a\bigg)$
$n_{KMnO_4}=\frac{15,8}{158}=0,1(mol)$
Chất rắn sau p/ứ là $K_2MnO_4,MnO_2$
Theo PT: $n_{K_2MnO_4}=n_{MnO_2}=0,05(mol)$
$\to m_{\rm chất\, rắn}=0,05.197+0,05.87=14,2(g)$
$b\bigg)$
Vì $H=80\%\to n_{KMnO_4(p/ứ)}=0,1.80\%=0,08(mol)$
$\to n_{KMnO_4(dư)}=0,02(mol)$
Chất rắn sau p/ứ là $KMnO_4(dư):0,02;K_2MnO_4:0,04;MnO_2:0,04$
$\to m_{\rm chất\, rắn}=0,02.158+0,04.197+0,04.87=14,52(g)$
$c\bigg)$
Bảo toàn KL có:
$m_{O_2}=m_{KMnO_4}-m_{CR}$
$\to m_{O_2}=15,8-14,68=1,12(g)\to n_{O_2}=0,035(mol)$
Theo PT: $n_{KMnO_4(p/ứ)}=2n_{O_2}=0,07(mol)$
$\to H=\dfrac{0,07}{0,1}.100\%=70\%$
\(n_S=\dfrac{32}{32}=1mol\\ S+O_2\xrightarrow[]{t^0}SO_2\\ n_{SO_2}=n_S=1mol\\ m_{SO_2\left(lt\right)}=1.64=64g\\ m_{SO_2\left(tt\right)}=64\cdot80:100=51,2g\)