Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng ĐLBTKL :
mAl + mO2 = mAl2O3
8,1 + 4,032 : 22,4 × 32 = 13,86 (g)
\(n_{O_2} = \dfrac{4,032}{22,4} = 0,18(mol)\\ n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ \dfrac{n_{Al}}{4} = 0,075 < \dfrac{n_{O_2}}{3} = 0,06\)
Suy ra: Al dư
Bảo toàn khối lượng :
\(m = m_{Al\ dư} + m_{Al_2O_3} = m_{Al\ dư} + m_{Al\ pư} + m_{O_2}=m_{Al\ ban\ đầu} + m_{O_2} = 8,1 + 0,18.32 = 13,86(gam)\)
\(n_{O_2}=\dfrac{20-15,2}{32}=0,15\left(mol\right)\)
=> V = 0,15.22,4 = 3,36 (l)
=> D
\(n_{H_2S\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2H_2S+3O_2-^{t^o}>2SO_2+2H_2O\)
tỉ lệ: 2 : 3 : 2 : 2
n(mol) 0,5---->0,75------>0,5------->0,5
\(m_{SO_2}=n\cdot M=0,5\cdot64=32\left(g\right)\)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
nMg = 5,76/24 = 0,24 (mol)
PTHH: 2Mg + O2 -> (t°) 2MgO
nMgO = 0,24 (mol)
mMgO = 0,24 . 40 = 9,6 (g)
nMg = 5,76 : 24 = 0,24 ( mol )
pthh : 2Mg+ O2 -t--> 2MgO
0,24->0,12-->0,24 (mol)
=> m = mMgO = 0,24 . 40 = 9,6 (g)
\(n_{Na}=\dfrac{3,68}{23}=0,16\left(mol\right)\\ PTHH:4Na+O_2\underrightarrow{t^o}2Na_2O\left(natri.oxit\right)\\ Theo.pt:n_{O_2}=\dfrac{1}{4}n_{Na}=\dfrac{1}{4}.0,16=0,04\left(mol\right)\\ V_{O_2}=0,04.22,4=0,896\left(l\right)\)
\(n_{Cu}=\dfrac{4}{80}=0.05\left(mol\right)\)
\(n_{Fe_2O_3}=\dfrac{4}{160}=0.025\left(mol\right)\)
\(n_{SO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(m_O=m_A-m_{Cu}-m_{Fe}-m_O=9.2-0.05\cdot64-0.025\cdot2\cdot56-0.1\cdot32=0\)
\(CT:Cu_xFe_yS_z\)
\(x:y:z=0.05:0.05:0.1=1:1:2\)
\(CT:CuFeS_2\)
\(n_{CH_4}=\dfrac{1,6}{16}=0,1\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,1--->0,2----->0,1---->0,2
\(\Rightarrow\left\{{}\begin{matrix}V=V_{CO_2}=0,1.22,4=2,24\left(l\right)\\m=m_{H_2O}=0,2.18=3,6\left(g\right)\end{matrix}\right.\)
\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(0.08......0.1......0.04\)
\(m_{rắn}=m_{P\left(dư\right)}+m_{P_2O_5}=\left(0.1-0.08\right)\cdot31+0.04\cdot142=6.3\left(g\right)\)