\(1)PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\\ n_{CaCO_3}=\dfrac{500.95\%}{100}=4,75(mol)\\ \Rightarrow n_{CaO}=4,75(mol)\\ \Rightarrow m_{CaO}=4,75.56=266(g)\\ \Rightarrow m_{CaO(tt)}=266.80\%=212,8(g)\\ m_{CaCO_3(k p/ứ)}=500.95\%.20\%=95(g)\\ \Rightarrow m_A=95+212,8=307,8(g)\\ 2)\%m_{CaO}=\dfrac{212,8}{307,8}.100\%=69,136\%\\ n_{CO_2}=n_{CaO}=4,75(mol)\\ \Rightarrow V_{CO_2}=4,75.22,4=106,4(l)\)

Giả sử có 100g đá

=> \(m_{CaCO_3}=\dfrac{100.80}{100}=80\left(g\right)\)

\(n_{CaCO_3}=\dfrac{80}{100}=0,8\left(mol\right)\)

Gọi số mol CaCO₃ phân hủy

PTHH: CaCO₃ --t^o--> CaO + CO₂

             a-------------->a--->a

=> m_Y = 100 - 44a (g)

=> m_CaO = 56a (g)

=> \(\dfrac{56a}{100-44a}.100\%=45,65\%\)

=> a = 0,6 (mol)

=> \(H=\dfrac{0,6}{0,8}.100\%=75\%\)

lấy m=100g 
=> mCaCO3=80 => n=0.8 
n pứ=x => mCaO=56x 
56x/(56x+(0.8-x)*100+20)=0.4565 
=> x=0.6 =|> H=0.6/0.8=0.75=75%

Đáp án A

Giả sử m=100(g)

Câu 7:

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Vì:\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow H_2SO_4dư,Znhết\\ a,n_{H_2}=n_{Zn}=0,3\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,n_{O\left(mất\right)}=n_{H_2O}=n_{H_2}=0,3\left(mol\right)\\ \Rightarrow m_{giảm}=m_{O\left(mất\right)}=0,3.16=4,8\left(g\right)\\ \Rightarrow m=4,8\left(g\right)\)

Câu 6:

- Giả sử có 1 mol hỗn hợp khí A. 

\(\Rightarrow\left\{{}\begin{matrix}n_{N_xO}=30\%.1=0,3\left(mol\right)\\n_{SO_2}=30\%.1=0,3\left(mol\right)\\n_{CO_2}=1-\left(0,3+0,3\right)=0,4\left(mol\right)\end{matrix}\right.\\ \Rightarrow m_A=0,3.\left(14x+16\right)+0,3.64+0,4.44=41,6+4,2x\left(g\right)\\ \%m_{N_xO}=19,651\%\\ \Leftrightarrow\dfrac{4,2x+4,8}{41,6+4,2x}.100\%=19,651\%\\ \Leftrightarrow x=1\\ \Rightarrow N_xO.là:NO\\ M_{hhA}=\dfrac{0,3.30+0,3.64+0,4.44}{1}=45,8\left(\dfrac{g}{mol}\right)\\ \Rightarrow d_{\dfrac{hhA}{H_2}}=\dfrac{45,8}{2}=22,9\)

Câu 4:

\(a,m_{CaCO_3\left(ban.đầu\right)}=95\%.m_{đá.vôi}=95\%.500=475\left(g\right)\\ \Rightarrow n_{CaCO_3\left(ban.đầu\right)}=\dfrac{475}{100}=4,75\left(mol\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ n_{CaO\left(LT\right)}=n_{CaCO_3\left(ban.đầu\right)}=4,75\left(mol\right)\\ Vì:H=80\%\Rightarrow n_{CaO\left(TT\right)}=4,75.80\%=3,8\left(mol\right)\\ \Rightarrow m_{rắnA}=m_{tạp.chất}+m_{CaCO_3\left(không.p.ứ\right)}+m_{CaO\left(TT\right)}\\ =25+\left(4,75-3,8\right).100+3,8.56=332,8\left(g\right)\\ b,\%m_{CaO}=\dfrac{3,8.56}{332,8}.100\approx63,942\%\\ n_B=n_{CO_2\left(TT\right)}=n_{CaO\left(TT\right)}=3,8\left(mol\right)\\ \Rightarrow V_{B\left(đktc\right)}=V_{CO_2\left(đktc\right)}=3,8.22,4=85,12\left(l\right)\)

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$a\bigg)$

$n_{KMnO_4}=\frac{15,8}{158}=0,1(mol)$

Chất rắn sau p/ứ là $K_2MnO_4,MnO_2$

Theo PT: $n_{K_2MnO_4}=n_{MnO_2}=0,05(mol)$

$\to m_{\rm chất\, rắn}=0,05.197+0,05.87=14,2(g)$

$b\bigg)$

Vì $H=80\%\to n_{KMnO_4(p/ứ)}=0,1.80\%=0,08(mol)$

$\to n_{KMnO_4(dư)}=0,02(mol)$

Chất rắn sau p/ứ là $KMnO_4(dư):0,02;K_2MnO_4:0,04;MnO_2:0,04$

$\to m_{\rm chất\, rắn}=0,02.158+0,04.197+0,04.87=14,52(g)$

$c\bigg)$

Bảo toàn KL có:

$m_{O_2}=m_{KMnO_4}-m_{CR}$

$\to m_{O_2}=15,8-14,68=1,12(g)\to n_{O_2}=0,035(mol)$

Theo PT: $n_{KMnO_4(p/ứ)}=2n_{O_2}=0,07(mol)$

$\to H=\dfrac{0,07}{0,1}.100\%=70\%$

1.

2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,1----------------0,05------0,05

n KMnO4=\(\dfrac{15,8}{158}\)=0,1 mol

=>n cr=0,05.197+0,05.87=14,2g

b)

=>n cr=(0,05.197+0,05.87).\(\dfrac{80}{100}\)=11,36g

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$a\bigg)$

$n_{KMnO_4}=\frac{15,8}{158}=0,1(mol)$

Chất rắn sau p/ứ là $K_2MnO_4,MnO_2$

Theo PT: $n_{K_2MnO_4}=n_{MnO_2}=0,1(mol)$

$\to m_{\rm chất\, rắn}=0,1.197+0,1.87=28,4(g)$

$b\bigg)$

Vì $H=80\%\to n_{KMnO_4(p/ứ)}=0,1.80\%=0,08(mol)$

$\to n_{KMnO_4(dư)}=0,02(mol)$

Chất rắn sau p/ứ là $KMnO_4(dư):0,02;K_2MnO_4:0,08;MnO_2:0,08$

$\to m_{\rm chất\, rắn}=0,02.158+0,08.197+0,08.87=25,88(g)$

$c\bigg)$

Bảo toàn KL có:

$m_{O_2}=m_{KMnO_4}-m_{CR}$

$\to m_{O_2}=15,8-14,68=1,12(g)\to n_{O_2}=0,035(mol)$

Theo PT: $n_{KMnO_4(p/ứ)}=2n_{O_2}=0,07(mol)$

$\to H=\dfrac{0,07}{0,1}.100\%=70\%$

\(a,n_{CuO}=\dfrac{59}{80}=0,7375\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2\uparrow+O_2\uparrow\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}n_{CuO}=0,36875\left(mol\right)\\n_{NO_2}=2n_{CuO}=1,475\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,36875\cdot22,4=8,26\left(l\right)\\V_{NO_2}=1,475\cdot22,4=33,04\left(l\right)\end{matrix}\right.\)

\(b,\text{Chất rắn thu đc là }CuO\text{ gồm có }Cu,O\\ \%_O=\dfrac{16}{80}\cdot100\%=20\%\\ \Rightarrow m_O=59\cdot20\%=11,8\left(g\right)\\ \Rightarrow m_{Cu}=59-11,8=47,2\left(g\right)\)