Đáp án + Giải thích các bước giải:

Câu 5:

$n_{Cu}=\dfrac{11,52}{64}=0,18(mol)$

$n_{ AgNO_3}=200.10^{-3}.1,5=0,3(mol)$

Phản ứng xảy ra:

$Cu+2AgNO_3 \to Cu(NO_3)_2+2Ag$

Nhận thấy $:\dfrac{0,18}{1} > \dfrac{0,3}{2}$ nên $Cu$ dư, tính theo $AgNO_3$

$n_{Ag}=n_{AgNO_3}=0,3(mol)$

$n_{Cu(dư)}=0,18-\dfrac{1}{2}.n_{Ag}=0,18-\dfrac{1}{2}.0,3=0,03(mol)$

$m=m_{Ag}+m_{Cu(dư)}=0,3.108+0,03.64=34,32(g)$

b.

$V_{Y}=V_{AgNO_3}=200(ml)$

$n_{Cu(NO_3)_2}=\dfrac{1}{2}.n_{AgNO_3}=\dfrac{1}{2}.0,3=0,15(mol)$

$[Cu(NO_3)_2]=\dfrac{0,15}{200.10^{-3}}0,75(M)$

Câu 6:

- Oxide:

$K_2O:$ potassium oxide.

$N_2O_5:$ dinitrogen pentaoxide.

- Acid:

$H_2SO_4:$ sulfuric acid.

- Base:

$Ca(OH)_2:$ calcium hydroxide.

- Muối:

$NaHCO_3:$ sodium hydrocarbonate.

$Mg(NO_3)_2:$ magnesium nitrate.

$Ba_3(PO_4)_2:$ barium phosphate.

$Fe_2(SO_4)_3:$ iron(III) sulfate.