Bài 19: Chuyển đổi giữa khối lượng, thể tích và lượng chất

19) \(n_{SO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

20) \(n_{NO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

22) \(V_{CO_2\left(đkc\right)}=n.24,79=0,3.24,79=7,437\left(l\right)\)

23) \(V_{O_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\)

24) \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{16}{64}=0,25\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\end{matrix}\right.\)

`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}=0,25+0,1=0,35\left(mol\right)\)

`=>` \(V_{hh\left(đkc\right)}=n.24,79=0,35.24,79=8,6765\left(l\right)\)

25) \(n_{CO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

`=>` \(m_{CO_2}=n.M=0,2.44=8,8\left(g\right)\)

26) \(n_{N_2}=\dfrac{m}{M}=\dfrac{28}{28}=1\left(mol\right)\)

`=>` \(V_{N_2\left(đkc\right)}=n.24,79=1.24,79=24,79\left(l\right)\)

27) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{22}{44}=0,5\left(mol\right)\)

`=>` \(V_{CO_2\left(đkc\right)}=n.24,79=0,5.24,79=12,395\left(l\right)\)

28) \(n_{NO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

`=>` \(m_{NO_2}=n.M=0,6.46=27,6\left(g\right)\)

29) \(n_{MgO}=\dfrac{m}{M}=\dfrac{24}{40}=0,6\left(mol\right)\)

`=>` \(\text{Số phân tử MgO}=n.6.10^{23}=0,6.6.10^{23}=3,6.10^{23}\left(\text{phân tử}\right)\)

a)

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{6,4}{32+16\cdot2}=0,1\left(mol\right)\\ V_{SO_2}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)

b)

\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ m_{CO_2}=n\cdot M=0,5\cdot\left(12+16\cdot2\right)=22\left(g\right)\)

a)
n_so2= \(\overline{\dfrac{m}{M}}\)=\(\dfrac{6,4}{64}\)=0,1 ( mol)
v_so2= n*22,4 = 0,1* 22,4=2,24(lít)

b) n_co2 = \(\dfrac{v}{24,79}\)= \(\dfrac{11,2}{22,4}\)= 0,5 (mol)

m_co2= n*m= 0,5 * 44= 22(gam)

xg r nè bn iu❤

`a) M_A = (25,6)/(0,4) = 64(g//mol)`

`b) M_B = (17,75)/(0,25) = 71(g//mol)`

\(\left\{{}\begin{matrix}n_{khí}=n_{O_2}+n_{NO_2}=0,25+0,35=0,6\left(mol\right)\\V_{khí}=22,4.n_{khí}=0,6.22,4=13,44\left(l\right)\end{matrix}\right.\)

$n_{H_2} = \dfrac{2}{2} = 1(mol)$
$n_{NH_3} = \dfrac{34}{17} = 2(mol)$
$\Rightarrow n_{hh} = 1 + 2 = 3(mol)$
Ta có : $n = \dfrac{PV}{RT} = \dfrac{1.V}{0,082(273 + 0)} = 3$
$\Rightarrow V = 67,158(lít)$

$n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : 

$n_{H_2} = n_{Fe} = 0,1(mol)$

$n_{HCl} = 2n_{Fe} = 0,2(mol)$

Suy ra : 

$V_{H_2} = 0,1.22,4 = 2,24(lít)$
$m_{HCl} = 0,2.36,5 = 7,3(gam)$

Bài 1 : 

$a) n_{C_2H_4} = \dfrac{2,24}{22,4} = 0,1(mol) ; n_{CO_2} = \dfrac{3,36}{22,4} = 0,15(mol) ; n_{N_2} = \dfrac{10,08}{22,4} = 0,45(mol)$
$m_{C_2H_4} = 0,1.28 = 2,8(gam)$
$m_{CO_2} = 0,15.44 = 6,6(gam)$
$m_{N_2} = 0,45.28 = 12,6(gam)$

b) $n_{CO_2} = \dfrac{16,8}{22,4} = 0,75(mol) ; n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)$
$n_{hỗn\ hợp} = 0,75 + 0,5 = 1,25(mol)$
$m_{hỗn\ hợp} = m_{CO_2} + m_{O_2} = 0,75.44 + 0,5.32 = 49(gam)$

Bài 2 : 

$a) V_{hỗn\ hợp} = (0,25 + 0,15).22,4 = 8,96(lít)$
$b) m_{hỗn\ hợp} = m_{SO_2} + m_{CO_2} = 0,25.64 + 0,15.44 = 22,6(gam)$

a, \(m_{Cu}=0,5.64=32\left(g\right)\)

b, \(m_{C_6H_{12}O_6}=0,8.180=144\left(g\right)\)

c, \(n_{C_4H_{10}}=\dfrac{7,392}{22,4}=0,33\left(mol\right)\)

\(\Rightarrow m_{C_4H_{10}}=0,33.58=19,14\left(g\right)\)

d, \(n_{CO_2}=\dfrac{21,12}{22,4}=\dfrac{33}{35}\left(mol\right)\)

\(\Rightarrow m_{CO_2}=\dfrac{33}{35}.44\approx41,49\left(g\right)\)

e, \(n_{Ca\left(NO_3\right)_2}=\dfrac{0,129.10^{25}}{6.10^{23}}=2,15\left(mol\right)\)

\(\Rightarrow m_{Ca\left(NO_3\right)_2}=2,15.164=352,6\left(g\right)\)