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\(m_{NaOH\left(A\right)}=20.5\%=1\left(g\right)\)
Trong B:
gọi x là khối lượng Na2O thêm vào , x>0 (g)
\(10\%=\dfrac{\dfrac{80}{62}x+1}{x+20}\)
\(\rightarrow x=0,84\left(g\right)\)
Vậy khối Na2O thêm vào dd A là 0,84 (g)
b, \(m_{KOH\left(A\right)}=2\%.20=0,4\left(g\right)\)
\(C\%_{KOH\left(B\right)}=\dfrac{0,4}{20+0,84}.100\%=1,92\%\)
Bài1:
a,Vì dd A là dd bazo nên làm cho quỳ tím đổi thành màu xanh
b,\(n_{Na_2O}=\dfrac{21,7}{62}=0,35\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,35 0,7
\(\Rightarrow C_{M_{ddNaOH}}=\dfrac{0,7}{0,4}=1,75M\)
Bài 2:
a,\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,15 0,3 0,15
⇒ a=mZn = 0,15.65 = 9,75 (g)
b,\(V_{HCl}=\dfrac{0,3}{1,5}=0,2\left(l\right)=200\left(ml\right)\)
a, \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)
PTHH: Na2O + H2O → 2NaOH
Mol: 0,05 0,1
⇒ ddA (NaOH) làm quỳ tím đổi màu xanh
b, \(C_{M_{ddNaOH}}=\dfrac{0,1}{0,5}=0,2M\)
c, \(m_{ddA}=3,1+1.500=503,1\left(g\right)\)
d, \(C\%_{ddNaOH}=\dfrac{0,2.40.100\%}{503,1}=1,59\%\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{193,8+6,2}\cdot100\%=4\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO4 còn dư, NaOH p/ứ hết
\(\Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
a) PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)
Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{20\%}=147\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{147}{1,14}\approx128,95\left(ml\right)\)
b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)=n_{FeSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\V_{H_2}=0,3\cdot24,76=7,428\left(l\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=163,2\left(g\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{163,2}\cdot100\%\approx27,94\%\)
\(m_{H_2SO_4\left(bđ\right)}\)=49%.200=98(g)
Gọi x là số mol SO3
=>a=\(m_{SO_3}\)=80x(g)
Ta có PTHH:
SO3+H2O->H2SO4
x.....................x..................(mol)
Theo PTHH:\(m_{H_2SO_4\left(pt\right)}\)=98x
=>\(m_{H_2SO_4}\)=98x+98(g)
Ta có:mdd(sau)=200+a=200+80x
=>\(C_{\%H_2SO_4\left(sau\right)}\)=\(\dfrac{98x+98}{200+80x}\).100%=60%
=>x=0,44(mol)
=>\(m_{SO_3}\)=80x=80.0,44=35,2(g)
2)\(n_{Na_2O}\)=62:62=1(mol)
\(n_{P_2O_5}\)=14,2:142=0,1(mol)
Ta có PTHH:
Na2O+H2O->2NaOH(1)
1...........................2................(mol)
P2O5+3H2O->2H3PO4(2)
0,1....................0,2.............(mol)
Theo PTHH(1);(2):
mNaOH=2.40=80(g)
\(m_{H_3PO_4}\)=0,2.98=19,6(g)
Ta có:mdd(sau)=62+14,2+500=576,2(g)
=>\(C_{\%NaOH}\)=\(\dfrac{80}{576,2}\).100%=13,9%
=>\(C_{\%H_3PO_4}\)=\(\dfrac{19,6}{576,2}\).100%=3,4%