11,2 l 

...đề đúng ra là "khối lượng ZnO thu được gấp..." chứ nhỉ

2Zn + O₂ \(\underrightarrow{to}\) 2ZnO (1)

2Mg + O₂ \(\underrightarrow{to}\) 2MgO (2)

Ta có: \(\dfrac{m_{ZnO}}{m_{MgO}}=\dfrac{2,025}{1}=\dfrac{81}{40}\)

\(\Rightarrow m_{ZnO}=12,1\div\left(81+40\right)\times81=8,1\left(g\right)\)

\(\Rightarrow m_{MgO}=12,1-8,1=4\left(g\right)\)

\(\Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

Theo PT1: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,1\times65=6,5\left(g\right)\)

Theo PT2: \(n_{Mg}=n_{MgO}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)

gọi x là khối lượng MgO (g), khối lượng ZnO là 2,025x (g)

ta có:

\(m_{mgO}+m_{ZnO}=12,1\Leftrightarrow x+2,025x=12,1\\ \Leftrightarrow3,025x=12,1\\ \Leftrightarrow x=\dfrac{12,1}{3,025}=4\left(g\right)\\ m_{MgO}=4\left(g\right)\Rightarrow m_{ZnO}=2,025\cdot4=8,1\left(g\right)\)

ta có PTHH(1): \(2Mg+O_2-t^0\rightarrow2MgO_{ }\)

theo gt:

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ theoPTHH:n_{O2}=2n_{MgO}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ n_{Mg}=n_{MgO}=0,1\left(mol\right)\Rightarrow m_{Mg}=o,1\cdot24=2,4\left(g\right)\)

PTHH(2):

\(2Zn+O_2-t^0\rightarrow2ZnO\\ theogt:n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ theoPTHH:n_{O2}=\dfrac{1}{2}n_{ZnO}=\dfrac{1}{2}0,1=0,05\left(mol\right)\\ n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1\cdot65=6,5\left(g\right)\)

từ PTHH(1) và (2) \(\Rightarrow n_{O2}=0,05+0,05=0,1\left(mol\right)\\ \Rightarrow V_{O2}=0,1\cdot22,4=2,24\left(l\right)\Rightarrow V_{kk}=2,24\cdot5=11,2\left(l\right)\)

Gọi n_Mg= x ; n_Zn= y (x , y >0)

PTHH :

2Mg + O₂\(\dfrac{t^o}{ }\)> 2MgO (1)

x ---->\(\dfrac{x}{2}\) ----->x

2Zn + O₂ \(\dfrac{t^o}{ }\)> 2ZnO (2)

y---->\(\dfrac{y}{2}\)------>y

Theo đề bài ta có :

40x + 81y = 12,1

và 81y = 2,025 . 40x

=> x = 0,1 ; y = 0,1

Theo pt (1) n_Mg=n_MgO= 0,1 mol

=> m_Mg = 2,4 g

Theo pt (2) n_Zn=n_ZnO=0,1 mol

=> m_Zn = 6,5 g

m_hh = 2,4+ 6,5 = 8,9 g

%Mg =\(\dfrac{2,4}{8,9}\) =26,97 %

%Zn = 100% - 26,97% = 73,03 %

V_O2=n_O2 . 22,4 = ......

@HUNG nguyen giúp mik với

\(m_{tăng}=m_{O_2}=7.2\left(g\right)\)

\(n_{O_2}=\dfrac{7.2}{32}=0.225\left(mol\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.225\cdot22.4=25.2\left(l\right)\)

\(Đặt:n_{Mg}a\left(mol\right),n_{Cu}=b\left(mol\right),n_{Al}=c\left(mol\right)\)

\(Mg+\dfrac{1}{2}O_2\underrightarrow{t^0}MgO\)

\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^0}CuO\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(TC:n_{O_2}=0.5a=0.5b=0.75c=\dfrac{0.225}{3}=0.075\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.15\\b=0.15\\c=0.1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0.15\cdot24=3.6\left(g\right)\\m_{Cu}=0.15\cdot64=9.6\left(g\right)\\m_{Al}=0.1\cdot27=2.7\left(g\right)\end{matrix}\right.\)

giúp tui với tui cần gấp

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\x_{Ca}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(m_{Mg}=0,4\cdot24=9,6g\)

   \(m_{Ca}=0,2\cdot40=8g\)

b)\(2Mg+O_2\underrightarrow{t^o}2MgO\)

   \(2Ca+O_2\underrightarrow{t^o}2CaO\)

Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)

\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

a) 

Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)

b)

PTHH: 2Ca + O₂ --t^o--> 2CaO 

            0,2-->0,1

             2Mg + O₂ --t^o--> 2MgO

            0,4--->0,2

=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)

\(V_{kk}=6,72.5=33,6\left(l\right)\)