Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,1-------------->0,1---->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
mdd sau pư = 2,3 + 197,8 - 0,05.2 = 200 (g)
=> \(C\%=\dfrac{0,1.40}{200}.100\%=2\%\)
\(V_{dd}=\dfrac{200}{1,08}=\dfrac{5000}{27}\left(ml\right)=\dfrac{5}{27}\left(l\right)\)
=> \(C_M=\dfrac{0,1}{\dfrac{5}{27}}=0,54M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.1.........................0.2\)
\(C_{M_{NaOH}}=\dfrac{0.2}{0.5}=0.4\left(M\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(0.2.............0.2\)
\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)
Chúc em học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{H_2SO_4}=0,3.0,75+0,3.0,25=0,3\left(mol\right)\\ V_{ddH_2SO_4}=300+300=600\left(ml\right)=0,6\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,6}=0,5M\\ m_{H_2SO_4}=0,3.98=29,4\left(g\right)\\ m_{ddH_2SO_4}=600.1,02=612\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{29,4}{612}.100\%=4,8\%\)
\(b,\) 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
\(\rightarrow H=\dfrac{0,1}{0,125}.100\%=80\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\uparrow\)
a) Ta có: \(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{KOH}=0,2\cdot56=11,2\left(g\right)\\m_{H_2}=0,1\cdot2=0,2 \left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_K+m_{H_2O}-m_{H_2}=400\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\dfrac{11,2}{400}\cdot100\%=2,8\%\)
b) Ta có: \(V_{dd\left(saup/ứ\right)}=\dfrac{400}{1,08}\approx370,37\left(ml\right)=0,37037\left(l\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,2}{0,37037}\approx0,54\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=\dfrac{m_1}{23}+m_2-\dfrac{m_1}{46}=\dfrac{m_1}{46}+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{m_1}{46}+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=m_1+m_2-\dfrac{m_1}{23}=\dfrac{22}{23}m_1+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{22}{23}m_1+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,3 0,3 0,3 0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
\(C_{M_{FeSO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂
200ml = 0,2 lít.
a) Số mol H₂: nH₂ = 6,72 ÷ 22,4 = 0,3 mol
Theo PTHH => Số mol Fe: nFe = 0,3 mol
=> Khối lượng Fe: mFe = 16,8g
b) Số mol H₂SO₄: nH₂SO₄ = 0,3 mol
Nồng độ mol dd: CM = 0,3 ÷ 0,2 = 1,5M
2Na + 2 H2O -> 2NaOH + H2
0,1 mol 0,1 mol
nNa = 2,3:23 = 0,1 mol
mdd sau phản ứng = ( 2,3 +197,8 ) - 0,1*2 = 199,9 g
Vdd sau phản ứng = mdd : D = 199,9 :1,08 \(\approx\)185,1 ml
nồng độ mol của dd là
CM = n : V = 0,1 : 0,1851 \(\approx\)0,54 M