Bài 23: Bài luyện tập 4

\(a.4NH_3+5O_2\rightarrow4NO+6H_2O\\ 4:5:4:6\)

\(b.S+2HNO_3\rightarrow H_2SO_4+2NO\\ 1:2:1:2\)

\(c.3Ba\left(NO_3\right)_2+Al_2\left(SO_4\right)_3\rightarrow3BaSO_4+2Al\left(NO_3\right)_3\\ 3:1:3:1\)

\(d.FeCl_3+3AgNO_3\rightarrow Fe\left(NO_3\right)_3+3AgCl\\ 1:3:1:3\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,1-->0,125------>0,05_______(mol)

=> \(\left\{{}\begin{matrix}m_{O_2}=0,125.32=4\left(g\right)\\m_{P_2O_5}=0,05.142=7,1\left(g\right)\end{matrix}\right.\)

Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

___0,1__0,125___0,05 (mol)

\(\Rightarrow m_{O_2}=0,125.32=4\left(g\right)\)

\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

Bạn tham khảo nhé!

\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\)

Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(n_{HCl(dư)}=0,6-0,2.2=0,2(mol)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ V_{H_2}=0,2.22,4=4,48(l)\)

\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)

Công thức khối lượng : 

\(m_P+m_{O_2}=m_{P_2O_5}\)

Khi đó : 

\(m_P=m_{P_2O_5}-m_P=14.2-6.2=8\left(g\right)\)

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

b) \(m_P+m_{O_2}=m_{P_2O_5}\)

c) áp dụng định luật bảo toàn khối lượng, ta có:

\(m_P+m_{O_2}=m_{P_2O_5}\)

\(\Rightarrow m_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)

vậy khối lượng oxi đã phản ứng là \(8g\)

a) \(n_{C_6H_{12}O_6}=\dfrac{7,2}{6}=1,2\left(mol\right)\)

\(\left\{{}\begin{matrix}m_H=1,2.12=14,4\left(g\right)\\m_O=1,2.6.16=115,2\left(g\right)\end{matrix}\right.\)

b) \(n_{C_{12}H_{22}O_{11}}=\dfrac{26,4}{22}=1,2\left(mol\right)\)

\(\left\{{}\begin{matrix}m_C=1,2.6.12=172,8\left(g\right)\\m_O=1,2.11.16=211,2\left(g\right)\end{matrix}\right.\)

Bài 3

\(a,PTHH:2KClO_3\rightarrow^{t^o}_{MnO_2}2KCl+3O_2\\ b,n_{KClO_3}=\dfrac{4,9}{122,5}=0,04\left(mol\right)\\ \Rightarrow n_{KCl}=0,04\left(mol\right)\\ \Rightarrow m_{KCl\left(p\text{/}ứ\right)}=0,04\cdot74,5=2,98\left(g\right)\\ \Rightarrow H\%=\dfrac{2,235}{2,98}\cdot100\%=75\%\\ c,n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,06\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,06\cdot22,4=1,344\left(l\right)\)

Bài 4:

\(a,n_{Cu\left(NO_3\right)_2}=\dfrac{15,04}{188}=0,08\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2+O_2\uparrow\\ \Rightarrow n_{CuO\left(p\text{/}ứ\right)}=0,08\left(mol\right)\\ \Rightarrow m_{CuO\left(p\text{/}ứ\right)}=0,08\cdot80=6,4\left(g\right)\\ \Rightarrow H\%=\dfrac{6,4}{8,56}\cdot100\%\approx75\%\\ b,n_{NO_2}=0,16\left(mol\right);n_{O_2}=0,04\left(mol\right)\\ \Rightarrow\overline{M_{hh}}=\dfrac{46\cdot0,16+32\cdot0,04}{0,16+0,04}=\dfrac{8,64}{0,2}=43,2\left(g/mol\right)\\ \Rightarrow d_{hh/H_2}=\dfrac{43,2}{2}=21,6\)

Câu 1:

\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ \Rightarrow n_{MgCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\ c,PTHH:H_2+CuO\rightarrow^{t^o}Cu+H_2O\\ \Rightarrow n_{Cu}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow m_{Cu}=0,2\cdot64=12,8\left(g\right)\)

Câu 2:

\(a,PTHH:3Fe+2O_2\rightarrow^{t^o}Fe_3O_4\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{Fe_3O_4}=\dfrac{13,92}{232}=0,06\left(mol\right)\\ \text{Vì }\dfrac{n_{Fe}}{3}< \dfrac{n_{Fe_3O_4}}{1}\Rightarrow Fe_3O_4\text{ dư}\\ \Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}\approx0,07\left(mol\right)\\ \Rightarrow V_{O_2}=0,07\cdot22,4=1,568\left(l\right)\)

\(a.M_{C_2H_6O}=12,2+2+16=46\left(đvC\right)\\ \%C=\dfrac{12.2}{46}.100=52,17\%\\ \%H=\dfrac{6}{46}.100=13,04\%\\ \%O=100-52,17-13,04=34,79\%\\ b.n_{CO_2}=\dfrac{6.6}{44}=0,15\left(mol\right)\\ BTNT\left(C\right):n_{C_2H_6O}.2=n_{CO_2}.1\\ \Rightarrow n_{C_2H_6O}=0,075\left(mol\right)\\ \Rightarrow m_{C_2H_6O}=0,075.46=3,45\left(g\right)\)