`a)`

$\rm m_{AgNO_3}=20 \times 12.75\%=2.55(gam)$

$\rm n_{AgNO_3}=\dfrac{2.55}{170}=0.015(mol)$

$\rm Cu+2AgNO_3\xrightarrow{} Cu(NO_3)_2+2Ag$

$\rm 0.0075 \leftarrow 0.015 \rightarrow 0.0075 \rightarrow 0.015 (mol)$

$\rm \rightarrow m_{Cu}=0.0075 \times 64=0.48(gam)$

$\rm \rightarrow m_{Cu(NO_3)_2}=0.0075 \times 188=1.41(gam)$

`b)`

$\rm n_{CO_2}=\dfrac{5.4538}{24.79}=0.22(mol)$

$\rm CO_2+2KOH \xrightarrow{} K_2CO_3+H_2O$

$\rm 0.22 \rightarrow 0.44 \rightarrow 0.22 \rightarrow 0.22 (mol)$

$\rm \rightarrow V_{dd}=\dfrac{0.44}{1.5}\approx0.29(L)=290(mL)$

`text[#RedSun]`.

a)

`Cu + 2AgNO_3 -> Cu(NO_3)_2 + 2Ag`

`n_{AgNO_3} = {20 . 12,75%}/{170} = 0,015` mol

`n_{Cu} = n_{Cu(NO_3)_2} = 1/2n_{AgNO_3} = 1/2 . 0,015= 0,0075` mol

`m_{Cu} = 0,0075 . 64 =0,48g`

`m_{Cu(NO_3)_2} = 0,0075 . 188 = 1,41g`

b)

`CO_2 + 2KOH -> K_2CO_3 + H_2O`

`n_{CO_2} = {5,4538}/{24,79} = 0,22` mol

`n_{KOH} = 2n_{CO_2} = 2 . 0,22 = 0,44` mol

`V_{dd KOH} = {0,44}/{1,5} ~~ 0,293l`