\(a,n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right);n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Vì:\dfrac{0,1}{1}>\dfrac{0,1}{2}\Rightarrow CaCO_3dư\\ n_{CO_2}=n_{CaCO_3\left(p.ứ\right)}=\dfrac{n_{HCl}}{2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow n_{CaCO_3\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow m_{CaCO_3\left(dư\right)}=0,05.100=5\left(g\right)\\ b,V_{CO_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

c) Muốn phản ứng xảy ra vừa đủ, chất nào thiếu ta cần thêm chất đó vào => Cần thêm HCl vào.

\(n_{HCl\left(cần\right)}=2.n_{CaCO_3}=2.0,1=0,2\left(mol\right)\\ \Rightarrow n_{HCl\left(thêm\right)}=0,2-0,1=0,1\left(mol\right)\\ \Rightarrow m_{HCl\left(thêm\right)}=0,1.36,5=3,65\left(g\right)\)

TK

https://hoidap247.com/cau-hoi/981113

nHCl=3.65/36.5=0.1(mol)

nCaCO3(dư)=5/100=0.05(mol)

nHCl=2⋅0.05=0.1(mol)

mHCl(ct)=0.1⋅36.5=3.65(g)

\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{3.65}{36.5}=0.1\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

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

\(0.1..............0.1\)

\(LTL:\dfrac{0.1}{1}>\dfrac{0.1}{2}\Rightarrow CaCO_3dư\)

\(m_{CaCO_3\left(dư\right)}=\left(0.1-0.05\right)\cdot100=5\left(g\right)\)

\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)

Để phản ứng xảy ra vừa đủ : 

\(n_{CaCO_3\left(dư\right)}=\dfrac{5}{100}=0.05\left(mol\right)\)

\(n_{HCl}=2\cdot0.05=0.1\left(mol\right)\)

\(m_{HCl\left(ct\right)}=0.1\cdot36.5=3.65\left(g\right)\)

tham khảo ở đây nha bạn:https://hoidap247.com/cau-hoi/981113

\(m_{HCl}=\dfrac{50\cdot3,65}{100}=1,825g\)

\(\Rightarrow n_{HCl}=0,05mol\)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

0,025         0,05         0,025                0,025

\(m=0,025\cdot100=2,5g\)

\(V=0,025\cdot22,4=0,56l\)

\(m_m=0,025\cdot111=2,775g\)

$a)$

$n_{HCl}=\dfrac{50.3,65\%}{36,5}=0,05(mol)$

$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$

Theo PT: $n_{CaCO_3}=0,025(mol)$

$\to m=0,025.100=2,5(g)$

$b)$

Theo PT: $n_{CO_2}=0,025(mol)$

$\to V=0,025.22,4=0,56(l)$

$c)$

Theo PT: $n_{CaCl_2}=0,025(mol)$

$\to m_{muối}=0,025.111=2,775(g)$

a+b) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{CO_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\\V_{CO_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{50\cdot40\%}{40}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) Tạo muối trung hòa, bazơ dư, tính theo CO₂

Bảo toàn Cacbon: \(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\)

\(n_{CuO}=\dfrac{6,4}{80}=0,08\left(mol\right)\)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

          0,08-->0,16----->0,08

=> m_HCl = 0,16.36,5 = 5,84 (g)

=> \(m_{dd.HCl}=\dfrac{5,84.100}{3,65}=160\left(g\right)\)

\(C\%_{CuCl_2}=\dfrac{0,08.135}{160+6,4}.100\%=6,49\%\)

nCO2=0,2mol

PTHH: 2HCl+CaCO3=>CaCl2+CO2+H2O

        0,4mol<-0,2mol<-0,2mol<-0,2mol->0,2mol

=> mHCl tham gia : 0,4.36,5=14,6g

=> C%HCl=14,6:100.100=14,6%

b)mCaCO3 tham gia : 0,2.100=20g

c) m muối thu được :0,2.111=22,2g

theo định luật btoan khối lượng ta có : m(CaCl2)=mHCl+mCaCO3-mCO2-mH2o

=100+20-0,2.44-0,1.18=109,4

=> C% muối: 22,2/109,4.100=20,29%

\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

a.

\(CaCO_3+2HNO_3\rightarrow Ca\left(NO_3\right)_2+H_2O+CO_2\)

0,1             0,2               0,1                        0,1

\(C\%_{dd.HNO_3}=\dfrac{0,2.63.100}{200}=6,3\%\)

b.

\(m_{dd.Ca\left(NO_3\right)_2}=10+200-0,1.44=205,6\left(g\right)\)

\(C\%_{dd.Ca\left(NO_3\right)_2}=\dfrac{0,1.164.100}{205,6}=7,98\%\)

a) \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(n_{Ba\left(OH\right)_2}=n_{BaO}=0,15\left(mol\right)\)

\(n_{BaCO_3}=\dfrac{23,64}{197}=0,12\left(mol\right)\)

Bảo toàn nguyên tố Ba => \(n_{Ba\left(HCO_3\right)_2}=0,15-0,12=0,03\left(mol\right)\)

Bảo toàn nguyên tố C: \(n_{CO_2}=0,12+0,03.2=0,18\left(mol\right)\)

=> \(V_{CO_2}=0,18.22,4=4,032\left(l\right)\)

b)Bảo toàn nguyên tố C : \(n_{CO_2}=n_{MgCO_3}+n_{CaCO_3}=a\left(mol\right)\)

Ta có : \(\dfrac{18,4}{100}< a< \dfrac{18,4}{84}\)

=> \(0,184< a< 0,22\)

\(n_{OH^-}=0,15.2=0,3\left(mol\right)\)

Lập T: \(\dfrac{0,3}{0,22}< T< \dfrac{0,3}{0,184}\)

=>\(1,36< T< 1,63\)

Do 1< \(1,36< T< 1,63\) <2

=> Phản ứng luôn tạo kết tủa