Tính C% em nha.
\(a.\)
\(n_{K_2O}=\dfrac{9.4}{94}=0.1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0.1.......................0.2\)
\(m_{KOH}=0.2\cdot56=11.2\left(g\right)\)
\(m_{dd}=9.4+100=109.4\left(g\right)\)
\(C\%_{KOH}=\dfrac{11.2}{109.4}\cdot100\%=10.23\%\)
\(b.\)
\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(0.1............................0.2\)
\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)
\(m_{dd}=14.2+200=214.2\left(g\right)\)
\(C\%_{H_3PO_4}=\dfrac{19.6}{214.2}\cdot100\%=9.15\%\)
\(c.\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.2.....................0.2..........0.1\)
\(m_{NaOH}=0.2\cdot40=8\left(g\right)\)
\(m_{dd}=4.6+120-0.1\cdot2=124.4\left(g\right)\)
\(C\%_{NaOH}=\dfrac{8}{124.4}\cdot100\%=6.43\%\)
a)\(n_{K_2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)
Ta có : 100g H2O ~ 100ml H2O
\(K_2O+H_2O\rightarrow2KOH\)
Theo PT nKOH = 2nK2O =0,2(mol)
=> CMKOH=\(\dfrac{0,2}{0,1}=2M\)
b) \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
Ta có : 200g H2O ~ 200ml H2O
Theo PT nH3PO4 = 2nP2O5 =0,2(mol)
=> \(CM_{H_3PO_4}=\dfrac{0,2}{0,2}=1M\)
c) \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
Ta có : 120g H2O ~ 120ml H2O
\(n_{Na}=\dfrac{4,6}{23}=0,2\) (mol)
Ta có : nNaOH=nNa=0,2(mol)
=> \(CM_{NaOH}=\dfrac{0,2}{0,12}=1,67M\)
