a, \(n_{P_2O_5}=\dfrac{21,3}{142}=0,15\left(mol\right)\)

PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

Theo PT: \(n_{H_3PO_4}=2n_{P_2O_5}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_3PO_4}=0,3.98=29,4\left(g\right)\)

b, m _{dd sau pư} = 21,3 + 300 = 321,3 (g)

\(\Rightarrow C\%_{H_3PO_4}=\dfrac{29,4}{321,3}.100\%\approx9,15\%\)

1) Gọi số mol P₂O₅ là a (mol)

PTHH: P₂O₅ + 3H₂O --> 2H₃PO₄

                a----------------->2a

\(m_{H_3PO_4\left(tổng\right)}=98.2a+\dfrac{10.200}{100}=196a+20\left(g\right)\)

m_{dd sau pư} = 142a + 200 (g)

=> \(C\%_{dd.sau.pư}=\dfrac{196a+20}{142a+200}.100\%=17,93\%\)

=> a = 0,093 (mol)

=> m_P2O5 = 0,093.142 = 13,206 (g)

2)

a) \(n_{O_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

                0,9<-----------0,45<----0,45<----0,45

=> \(m_{KMnO_4\left(Pư\right)}=0,9.158=142,2\left(g\right)\)

=> \(m_{KMnO_4\left(tt\right)}=\dfrac{142,2.100}{80}=177,75\left(g\right)\)

=> \(m=\dfrac{177,75.100}{90}=197,5\left(g\right)\)

b) 

X \(\left\{{}\begin{matrix}m_{K_2MnO_4}=0,45.197=88,65\left(g\right)\\m_{MnO_2}=0,45.87=39,15\left(g\right)\\m_{KMnO_4}=177,75-142,2=35,55\left(g\right)\\m_{tạp.chất}=197,5.10\%=19,75\left(g\right)\end{matrix}\right.\)

giup mk voi

Bài 1:

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Na}=n_{Na_2O}=0,2.2=0,4\left(mol\right)\\ a.m_{Na}=0,4.23=9,2\left(g\right)\\ b.C_{MddA}=\dfrac{0,4}{0,5}=0,8\left(M\right)\\ C\%_{ddA}=\dfrac{0,4.40}{500.1,2}.100\approx2,667\%\)

a)

\(n_{NO} = 0,01(mol)\)

Bảo toàn electron : \(2n_{Cu} = 3n_{NO}\)

\(\Rightarrow n_{Cu} = 0,015(mol)\)

Vậy :

\(\%m_{Cu} = \dfrac{0,01.64}{1,98}.100\% = 32,32\%\\ \%m_{Al_2O_3} = 100\% - 32,32\% = 67,68\%\)

b)

\(n_{Al_2O_3} = \dfrac{1,98-0,015.64}{102} = 0,01(mol)\)

\(n_{OH^-} = n_{NaOH} + 2n_{Ba(OH)_2} = 0,1.1 + 0,5.0,1.2 = 0,2(mol)\)

Ta có :

\(n_{Cu(NO_3)_2} = n_{Cu} = 0,015(mol)\\ n_{Al(NO_3)_3} = 2n_{Al_2O_3} = 0,02(mol)\)

Mà : \(n_{OH^-} = 4n_{Al(NO_3)_3} + 2n_{Cu(NO_3)_2} + n_{HNO_3\ dư}\)

\(\Rightarrow n_{HNO_3\ dư} = 0,09(mol)\)

Bảo toàn nguyên tố với N :

\(n_{HNO_3\ pư} = 3n_{Al(NO_3)_3} + 2n_{Cu(NO_3)_2} = 0,06(mol)\)

Vậy, \(V = \dfrac{0,09 + 0,06}{1} = 0,15(lít)\)

\(Đặt:n_{Fe}=a\left(mol\right);n_S=b\left(mol\right)\left(a,b>0\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ \Rightarrow\left\{{}\begin{matrix}56a+32b=20\\\dfrac{44,8}{3}a+22,4b=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\\ a,\%m_{Fe}=\dfrac{0,3.56}{20}.100\%=84\%\\ \%m_S=100\%-84\%=16\%\\ b,m_{sp}=m_{hhFe,S}+m_{O_2}=20+0,3.32=29,6\left(g\right)\)

a) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\) (mol)

Gọi x,y lần lượt là số mol của Fe và S.(x,y>0)

Theo bài ra, ta có: 56x+32y=20  (I)

PTHH:          \(3Fe\)  + \(2O_2\)\(\underrightarrow{t^o}\)\(Fe_3O_4\).

Theo pt:       x\(\rightarrow\)      \(\dfrac{2}{3}x\)       \(\dfrac{1}{3}x\)          (mol)   (1)

PTHH:         \(S+O_2\)  \(\underrightarrow{t^o}\)  \(SO_2\).

Theo pt:       y \(\rightarrow\) y             y            (mol)   (2)

Từ (1) và (2) \(\Rightarrow\) \(\sum n_{O_2}=\dfrac{2}{3}x+y=0,3\) (mol)  (II)

Giải hệ phương trình (I) và (II) ta được \(\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\)

\(\Rightarrow\) \(m_{Fe}=0,3.56=16,8\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{16,8}{20}.100\%=84\%\)

\(\Rightarrow\%m_S=100\%-84\%=16\%\)

b) \(m_{Fe_3O_4}=0,1.\left(3.56+4.16\right)=23,2\left(g\right)\)

\(m_{SO_2}=0,1.\left(32+2.16\right)=6,4\left(g\right)\)

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)$
$m_{ZnCl_2} = 0,1.136 = 13,6(gam)$
b)

$n_{HCl} = 2n_{Zn} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M$

c)

CuO + H_2 \to Cu + H_2O$

$n_{CuO} = 0,125(mol) > n_{H_2} \to $ CuO$ dư

$n_{Cu} = n_{CuO\ pư} = n_{H_2} = 0,1(mol)$

$n_{CuO\ dư} = 0,125 - 0,1 = 0,025(mol)$

$\%m_{Cu} = \dfrac{0,1.64}{0,1.64 + 0,025.80}.100\% = 76,2\%$
$\%m_{CuO} = 23,8\%$

)

nHCl=2nZn=0,2(mol)⇒CMHCl=0,20,1=2MnHCl=2nZn=0,2(mol)⇒CMHCl=0,20,1=2M

c)

CuO + H_2 \to Cu + H_2O$

nCuO=0,125(mol)>nH2→nCuO=0,125(mol)>nH2→ CuO$ dư

nCu=nCuO pư=nH2=0,1(mol)nCu=nCuO pư=nH2=0,1(mol)

nCuO dư=0,125−0,1=0,025(mol)nCuO dư=0,125−0,1=0,025(mol)