\(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(1.........2............................1\)

\(V_{H_2}=1\cdot22.4=22.4\left(l\right)\)

\(m_{dd_{HCl}}=\dfrac{2\cdot36.5}{7.3\%}=1000\left(g\right)\)

\(n_{BaSO_4}=\frac{m}{M}=\frac{32,62}{233}=0,14mol\)

PTHH:

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

0,14 0,14 0,14 0,28 (mol)

Gọi \(V_{ddH_2SO_4}\)cần thêm là x

\(n_{H_2SO_4}=\frac{m}{M}=\frac{98}{98}=1mol\)

\(C^{\left(A\right)}_{M_{H_2SO_4}}=\frac{1}{1}=1M\)

\(n^{\left(A\right)}_{H2SO4}=C_M.V=1.x=xmol\)

\(n_{H2SO4}=C_M.V=2.0,4=0,8mol\)

\(C_{MX}=\frac{n}{V}=\frac{0,8+x}{0,4+x}\left(M\right)\)

\(n_X=C_{MX}.V\)

\(\Leftrightarrow0,14=\frac{0,8+x}{0,4+x}.0,1\)

\(\Leftrightarrow\frac{0,14}{0,1}=\frac{0,8+x}{0,4+x}\)

⇔0,08+0,1x=0,56+0,14x

⇔x=0,6(l)

Vậy cần thêm 0,6 l dung dịch

4,48l khi o2

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)

a) 4P + 5O₂ \(\underrightarrow{to}\) 2P₂O₅ (1)

b) Theo PT1: \(n_P=\dfrac{4}{5}n_{O_2}\)

theo bài: \(n_P=n_{O_2}\)

Vì \(1>\dfrac{4}{5}\) ⇒ P dư

Theo PT1: \(n_Ppư=\dfrac{4}{5}n_{O_2}=\dfrac{4}{5}\times0,4=0,32\left(mol\right)\)

\(\Rightarrow n_Pdư=0,4-0,32=0,08\left(mol\right)\)

\(\Rightarrow m_Pdư=0,08\times31=2,48\left(g\right)\)

c) Theo PT1: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{2}{5}\times0,4=0,16\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,16\times142=22,72\left(g\right)\)

\(\Rightarrow m_{sp}=m_Pdư+m_{P_2O_5}=2,48+22,72=25,2\left(g\right)\)

d) 2KMnO₄ \(\underrightarrow{to}\) K₂MnO₄ + MnO₂ + O₂ (2)

Theo PT2: \(n_{KMnO_4}=2n_{O_2}=2\times0,4=0,8\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,8\times158=126,4\left(g\right)\)

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)
a. PTHH: \(4P+5O_2-t^o->2P_2O_5\)
Theo PTHH và đề bài ta có tỉ lệ:
\(\dfrac{0,4}{4}=0,1>\dfrac{0,4}{5}=0,08\)
b. => P dư. \(O_2\) hết => tính theo \(n_{O_2}\)
Theo PT ta có: \(n_{P\left(pư\right)}=\dfrac{0,4.4}{5}=0,32\left(mol\right)\)
=> \(n_{P\left(dư\right)}=0,4-0,32=0,08\left(mol\right)\)
=> \(m_{P\left(dư\right)}=0,08.31=2,48\left(g\right)\)
c. Theo PT ta có: \(n_{P2O5}=\dfrac{0,4.2}{5}=0,16\left(mol\right)\)
=> \(m_{P2O5}=0,16.142=22,72\left(g\right)\)
=> \(m_{sảnphẩmthuđược}=m_{P\left(dư\right)}+m_{P_2O_5}=2,48+22,72=25,2\left(g\right)\)
d. PTHH: \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\uparrow\)
Ta có: \(n_{O_2}=0,4\left(mol\right)\)
Theo PT ta có: \(n_{KMnO_4}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
=> \(m_{KMnO_4\left(cầndùng\right)}=0,8.158=126,4\left(g\right)\)

bài mấy rứa

Đề này khó hiểu đó, đã khó hiểu mà bạn còn đánh ko dấu, mình giúp ít thôi.

PTHH: C + O₂ -t^o-> CO₂

Ta có:

\(n_C=\frac{3}{12}=0,25\left(mol\right)\)

Theo PTHH và đề bài, ta có:

\(n_{CO_2}=n_C=0,25\left(mol\right)\\ =>V_{CO_2\left(đktc\right)}=22,4.0,25=5,6\left(l\right)\)