Đáp án + Giải thích các bước giải:

Câu 11: C

Phản ứng xảy ra:

$C(s)+O_2(g)\xrightarrow{t^o}CO_2(g)$

$\xrightarrow[]{\text{BTKL}}m_{C}+m_{O_2}=m_{CO_2}\\\to100+48=m_{CO_2}\\\to m_{CO_2}=148(g)$

Câu 12: B

$n_{Al}=\dfrac{6,48}{27}=0,24(mol)$

Phản ứng xảy ra:

$2Al(s)+6HCl(aq) \to 2AlCl_3(aq)+3H_2(g)$

$n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,24=0,36(mol)$

$V_{H_2}(đkc)=0,36.24,79=8,9244(l)$

Câu 13: D

$n_{P}=\dfrac{1,24}{31}=0,04(mol)$

$n_{O_2}=\dfrac{1,92}{32}=0,06(mol)$

Phản ứng xảy ra:

$4P(s)+5O_2(g)\xrightarrow[]{t^o}2P_2O_5(s)$

Xét thấy $:\dfrac{0,04}{4} < \dfrac{0,06}{5}$ nên $O_2$ dư. Tính theo $P$

$n_{O_2(dư)}=0,06-\dfrac{5}{4}.n_{P}=0,06-\dfrac{5}{4}.0,04=0,01(mol)$

$m_{O_2(dư)}=0,01.32=0,32(g)$

Câu 14: B

$S_{NaCl}=\dfrac{17}{50}.100=34(g/100g H_2O)$

Câu 15: D

$n_{CuSO_4}=100.10^{-3}.0,5=0,05(mol)$

$m_{CuSO_4}=0,05.160=8(g)$

Câu 11C

`C+O_2 overset[t^@][->] CO_2`

BTKL

`m_C+m_[O_2]=m_[CO_2]`

`=> 100+48=m_[CO_2]`

`=> m_[CO_2]=148(g)`

Câu 12B

`2Al+6HCl->2AlCl_3+3H_2`

`n_[Al]=[6.48]/27=0.24(mol)`

`PT: n_[H_2]=3/2n_[Al]=0.36(mol)`

`V_[H_2]=0.36 xx 24.79=8.9244(L)`

Câu 13D

`4P+5O_2 overset[t^@][->] 2P_2O_5`

`n_P=[1.24]/31=0.04(mol)`

`n_[O_2]=[1.92]/32=0.06(mol)`

Xét tỉ lệ

`[n_P]/4=[0.04]/4=0.01`

`[n_[O_2]]/5=[0.06]/5=0.012`

`=> P` hết `O_2` dư

`n_[O_2] (du)=0.06-[0.04*5]/4=0.01(mol)`

`m_[O_2]=0.01 xx 32=0.32(g)`

Câu 14B

`S_[KCl] (20^@C)=17/50*100=34(g)`

Câu 15D

`n_[CuSO_4]=0.5 * 0.1=0.05(mol)`

`m_[CuSO_4]=0.05 xx 160=8(g)`

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