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\(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
\(P_2O_5+3H_2O\xrightarrow[]{}2H_3PO_4\)
0,05 → 0,15 → 0,1
\(\Rightarrow m_{H_3PO_4}=0,1\cdot98=9,8\left(g\right)\)
\(\Rightarrow m_{H_2O}\left(\text{pư}\right)=0,15\cdot18=2,7\left(g\right)\)
\(\Rightarrow m_{H_2O}\left(\text{dm}\right)=100-2,7=97,3\left(g\right)\)
\(\Rightarrow m_{H_3PO_4}\left(\text{dd}\right)=m_{H_3PO_4}+m_{H_2O}\left(\text{dm}\right)=9,8+97,3=107,1\left(g\right)\)
\(\Rightarrow C\%=\dfrac{m_{H_3PO_4}}{m_{H_3PO_4}\left(\text{dd}\right)}\cdot100\%=\dfrac{9,8}{107,1}\cdot100\%\approx9,15\%\)
\(n_{P_2O_5}=\dfrac{99,4}{142}=0,7\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
0,7 2,1 1,4
a, \(m_{H_3PO_4}=1,4.98=137,2\left(g\right)\)
\(m_{ddH_3PO_4}=99,4+500=599,4\left(g\right)\)
Kl nước trong dd A :
\(m_{H_2O}=599,4-137,2=462,2\left(g\right)\)
\(b,C\%_{H_3PO_4}=\dfrac{137,2}{599,4}.100\%\approx22,89\%\)
\(c,C_M=\dfrac{n}{V}=\dfrac{1,4}{0,5}=2,8M\)
a/
\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,15 0,3 (mol)
\(m_{NaOH}=0,3.40=12\left(g\right)\)
\(m_A=90,7+9,3=100\left(g\right)\)
\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)
b/
m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)
\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)
\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)
bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)
pư: 0,3 0,15 0,15 0,15 (mol)
dư: 0 \(\dfrac{23}{380}\) (mol)
\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)
\(m_C=100+200-13,5=286,5\left(g\right)\)
\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)
\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)
\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)
\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)
Zn + 2HCl ---> ZnCl2 + H2
0.2-→0.4------→0.2--→0.2 (mol)
nZn = 11,2\56 = 0.2(mol)
mZnCl2 = n*M = 0.2*127 = 25.4(g)
VH2(đktc) = n*22.4 = 0.2*22.4 = 4.48(l)
mHCl = n*M = 0.4*36.5 = 14.6(g)
C% =14,6\146*100% = 10(%)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{11,2}{65}=0,17\left(mol\right)\\ TheoPT:n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,17\left(mol\right)\\ m_{ZnCl_2}=0,17.136=23,12\left(g\right)\\ V_{H_2}=0,17.22,4=3,808\left(l\right)\\ c.n_{HCl}=2n_{Zn}=0,34\left(mol\right)\\ C\%_{HCl}=\dfrac{0,34.36,5}{146}.100=8,5\%\)
a) Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH: Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
0,1---->0,3------->0,1
=> m = 0,1.342 = 34,2 (g)
c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)
\(a,n_{H_2SO_4}=0,3.0,75+0,3.0,25=0,3\left(mol\right)\\ V_{ddH_2SO_4}=300+300=600\left(ml\right)=0,6\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,6}=0,5M\\ m_{H_2SO_4}=0,3.98=29,4\left(g\right)\\ m_{ddH_2SO_4}=600.1,02=612\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{29,4}{612}.100\%=4,8\%\)
\(b,\) Đặt kim loại M có hoá trị n (n ∈ N*)
PTHH: \(2M+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nH_2\uparrow\)
\(\dfrac{0,6}{n}\)<---0,3--------------------------->0,3
\(\rightarrow M_M=\dfrac{5,4}{\dfrac{0,6}{n}}=9n\left(g\text{/}mol\right)\)
Vì n là hoá trị của M nên ta xét bảng
| \(n\) | \(1\) | \(2\) | \(3\) |
| \(M_M\) | \(9\) | \(18\) | \(27\) |
| \(Loại\) | \(Loại\) | \(Al\) |
Vậy M là Al
\(c,n_{KClO_3}=\dfrac{15,3125}{122,5}=0,125\left(mol\right)\)
PTHH:
\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)
0,3-->0,15
\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\uparrow\)
0,1<---------------------0,15
\(\rightarrow H=\dfrac{0,1}{0,125}.100\%=80\%\)
`a)PTHH:`
`Mg + H_2 SO_4 -> MgSO_4 + H_2`
`0,2` `0,2` `0,2` `(mol)`
`n_[Mg]=[4,8]/24=0,2(mol)`
`b)m_[MgSO_4]=0,2.120=24(g)`
`c)C%_[MgSO_4]=24/[4,8+50-0,2.2].100~~44,12%`
Câu 6:
Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PT: \(P_2O_5+6KOH\rightarrow2K_3PO_4+3H_2O\)
____0,05____0,3_______0,1 (mol)
Ta có: m dd sau pư = 7,1 + 100 = 107,1 (g)
\(C\%_{K_3PO_4}=\dfrac{0,1.212}{107,1}.100\%\approx19,8\%\)
\(m_{KOH}=0,3.56=16,8\left(g\right)\)
\(\Rightarrow x=\dfrac{16,8}{100}.100\%=16,8\%\)
Bạn tham khảo nhé!