\(S_{Na_2CO_3}=\dfrac{53}{250}.100=21,2\) 
\(C\%=\dfrac{53}{250+53}.100\%=17,5\%\)

nK=0,5(mol)

PTHH: K + H2O -> KOH + 1/2 H2

nKOH=nK=0,5(mol) => mKOH=0,5. 56=28(g)

mddKOH=mK+mH2O-mH2=19,5+ 261- 0,25 x 2= 280(g)

=>C%ddKOH=(28/280).100=10%

C%dd=mct/mdd.100%=19,5/19,5+261.100%≈6,95%

\(n_K=\dfrac{19.5}{39}=0.5\left(mol\right)\)

\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)

\(0.5................0.5............0.25\)

\(m_{KOH}=0.5\cdot56=28\left(g\right)\)

\(m_{dd}=19.5+261-0.25\cdot2=280\left(g\right)\)

\(C\%_{KOH}=\dfrac{28}{280}\cdot100\%=10\%\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(n_{SO_3}=\dfrac{12}{80}=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{SO_3}=0,15\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,15.98}{12+100}.100\%=13,125\%\)

GIÚP  dì ạ ?

bài đâu

ko gửi ảnh ak-.-

\(C\%_{KCl}=\dfrac{34}{100+34}\cdot100\%=25.37\%\)

Ta có: \(n_{SO_3}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(a,PTHH:SO_3+H_2O--->H_2SO_4\left(1\right)\)

Ta lại có: \(m_{dd_A}=0,25.80+100=120\left(g\right)\)

Theo PT₍₁₎: \(n_{H_2SO_4}=n_{SO_3}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow C_{\%_A}=\dfrac{24,5}{120}.100\%=20,42\%\)

\(b.PTHH:2KOH+H_2SO_4--->K_2SO_4+2H_2O\left(2\right)\)

Theo PT₍₂₎: \(n_{KOH}=2.n_{H_2SO_4}=2.0,25=0,5\left(mol\right)\)

\(\Rightarrow V_{dd_{KOH}}=\dfrac{0,5}{2}=0,25\left(lít\right)\)

nSO3=5,622,4=0,25(mol)nSO3=5,622,4=0,25(mol)

$a,PTHH:S{}^{}+{}^{}O--->{}^{}S{}^{}\left(1\right)$

Ta lại có: mddA=0,25.80+100=120(g)mddA=0,25.80+100=120(g)

Theo PT₍₁₎: nH2SO4=nSO3=0,25(mol)nH2SO4=nSO3=0,25(mol)

⇒mH2SO

$n_{Na_2O} = \dfrac{18,6}{62} = 0,3(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,6(mol)$
$C\%_{NaOH} = \dfrac{0,6.40}{200}.100\% = 12\%$

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{18,6}{62}=0,6\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,6\cdot40}{18,2+200}\cdot100\%\approx11\%\)

a) Ta có: \(C\%_{NaCl}=\dfrac{60}{60+1250}\cdot100\%\approx4,58\%\)

b) Ta có: \(C_{M_{Na_2CO_3}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

Bài 9:

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.100}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,4-2.0,1=0,2\left(mol\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)

Bài 10:

\(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ n_{Ca\left(OH\right)_2}=n_{CaO}=0,1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=0,1.74=7,4\left(g\right)\\ m_{ddCa\left(OH\right)_2}=m_{CaO}+m_{H_2O}=5,6+500.1=505,6\left(g\right)\\ a,C\%_{ddCa\left(OH\right)_2}=\dfrac{7,4}{505,6}.100\approx1,464\%\\ b,V_{ddCa\left(OH\right)_2}=V_{H_2O}=500\left(ml\right)=0,5\left(l\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)