Fe+2HCl->FeCl₂+H₂

0,1---------------------0,1

2H2+O₂-to>2H₂O

0,1----0,05 mol

0,1--0,1

n _Fe=\(\dfrac{5,6}{56}\)=0,1 mol

=>V_H2=0,1.22,4=2,24l

=>m_kk=0,05.29=1,45l

Không đủ dữ kiện tính $m_{kk}$ mà chỉ tính được $V_{kk}$

$a)PTHH:Fe+2HCl\to FeCl_2+H_2\uparrow$

$\to$ Phản ứng trao đổi

$b)n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$

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

$\Rightarrow V_{H_2(đktc)}=0,1.22,4=2,24(lít)$

$c)PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{O_2}=\dfrac{1}{2}n_{H_2}=0,05(mol)$

$\Rightarrow V_{O_2(đktc)}=0,05.22,4=1,12(lít)$

Mà $V_{O_2}$ chiếm $20\%$ của $V_{kk}$

$\Rightarrow V_{kk(đktc)}=\dfrac{1,12}{20\%}=5,6(lít)$

Bài 2:

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ n_{Fe}=n_{H_2}=0,45\left(mol\right);n_{HCl}=2.0,45=0,9\left(mol\right)\\ a,m_{Fe}=0,45.56=25,2\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,9}{0,15}=6\left(M\right)\)

\(n_{H_2SO_4}=\dfrac{980}{98}=10\left(mol\right)\)

\(2NaHCO_3+H_2SO_4\rightarrow Na_2SO_4+2CO_2+2H_2O\)

\(..............10....................20\)

\(V_{CO_2}=20\cdot22.4=448\left(l\right)\)

Đáp án : B

$n_{NaHCO_3}.84 + n_{MgCO_3}.84 = 16,8$
$n_{NaHCO_3} + n_{MgCO_3} = \dfrac{16,8}{84} = 0,2(mol)$
$n_{CO_2} = n_{NaHCO_3} + n_{MgCO_3} = 0,2(mol)$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,2(mol)$
$m_{CaCO_3} = 0,2.100 = 20(gam)$

\(n_{CaCO_3}=\dfrac{7,5}{100}=0,075\left(mol\right)\)

=> n_C = 0,075 (mol)

Có \(n_{CO_2}=n_C=0,075\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{4,2-0,075.44}{18}=0,05\left(mol\right)\)

=> n_H = 0,1 (mol)

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Bảo toàn O: \(n_{O\left(A\right)}=0,075.2+0,05-0,1.2=0\left(mol\right)\)

=> A chứa C, H

m_A = m_C + m_H = 0,075.12 + 0,1.1 = 1 (g)

\(m_{tăng}=m_{H_2O}+m_{CO_2}=4,2\left(g\right)\\ n_{CaCO_3}=\dfrac{7,5}{100}=0,075\left(mol\right)\)

PTHH: Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                            0,075        0,075

\(\rightarrow m_{CO_2}=0,075.44=3,3\left(g\right)\\ \rightarrow m_{H_2O}=4,2-3,3=0,9\left(g\right)\\ \rightarrow n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\\ \rightarrow n_{O\left(sau.pư\right)}=0,05+0,075.2=0,1\left(mol\right)\\ n_{O\left(trong.O_2\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}n_C=0,075\left(mol\right)\\n_H=0,05.2=0,1\left(mol\right)\\n_O=0,1-0,1=0\left(mol\right)\end{matrix}\right.\)

=> m_A = 0,075.12 + 0,1.1 + 0 = 1 (g)