Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl -> ZnCl₂ + H₂ (1)

a) Ta có: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

b) PTHH: 3H₂ + Fe₂O₃ -> 2Fe + 3H₂O (2)

Ta có: \(\left\{{}\begin{matrix}n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,3\left(mol\right)\\n_{Fe_2O_3\left(2\right)}=\dfrac{19,2}{160}=0,12\left(mol\right)\end{matrix}\right.\)

Ta có: \(\dfrac{0,3}{3}< \dfrac{0,12}{1}\)

=> Fe₂O₃ dư, H₂ hết nên tính theo H₂.

=> \(n_{Fe}=\dfrac{2.0,3}{3}=0,2\left(mol\right)\\ =>m_{Fe}=0,2.56=11,2\left(g\right)\)

a) PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Ta có: n_Zn = \(\dfrac{19,5}{65}\) = 0,3 mol'

Theo PT: n_Zn = n_H2 = 0,3 mol

=> V_H2 = 0,3 x 22,4 = 6,72l

b) PTHH: Fe₂O₃ + 3H₂ --> 2Fe + 3H₂O

\(n_{Fe_2O_3}\) = \(\dfrac{19,2}{160}\) = 0,12 mol

Vì \(\dfrac{0,12}{1}>\dfrac{0,3}{3}\) => Fe₂O₃ dư

Cứ 3 mol H₂ --> 2 mol Fe

0,3 mol --> 0,2 mol'

=> m_Fe = 0,2 x 56 = 11,2g

1.

a)

\(n_{KNO_3}=2.0,5=1\left(mol\right)\) => \(m_{KNO_3}=1.101=101\left(g\right)\)

b)

\(n_{CaCl_2}=0,25.0,1=0,025\left(mol\right)\)

=> \(m_{CaCl_2}=0,025.111=2,775\left(g\right)\)

2.

Nước tác dụng vs oxit axit --> dd axit

Nước tác dụng vs oxit bazơ-->dd bazơ

Nước tác dụng vs kim loại -->dd bazơ+hidro (những kloại có thể tác dụng Na, Ca , Ba ,K)

Bn tự vt PTHH đc ko ?

\(SO_2+H_2O-->H_2SO_3\)

\(Na_2O+H_2O-->2NaOH\)

\(Na+2H_2O-->NaOH+H_2\)

3.

a) \(C\%_{KCl}=\dfrac{20}{600}.100=3,34\%\)

b) \(C\%_{NaCl}=\dfrac{15}{45}.100=33,34\%\)

\(C\%_{KCl}=\dfrac{20}{600}\cdot100\%=3.33\%\)

\(C\%_{K_2SO_4}=\dfrac{75}{1500}\cdot100\%=5\%\)

\(C\%_{NaCl}=\dfrac{15}{15+45}\cdot100\%=25\%\)

\(n_{HCl}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

\(m_{dd_{HCl}}=7.3+500=507.3\left(g\right)\)

\(C\%_{HCl}=\dfrac{7.3}{507.3}\cdot100\%=1.44\%\)

a) +) Ta có: \(V_{ddKNO_3}=500\left(ml\right)=0,5\left(l\right)\\ =>n_{KNO_3}=0,5.2=1\left(mol\right)\\ =>m_{KNO_3}=1.101=101\left(g\right)\)

b) +) Ta có: \(V_{ddCaCl_2}=250\left(ml\right)=0,25\left(l\right)\\ =>n_{ddCaCl_2}=0,25.0,1=0,025\left(mol\right)\\ =>m_{CaCl_2}=111.0,025=2,775\left(g\right)\)

a) Đổi 500 ml =0,5l

\(n_{KNO_3}\) = 0,5 . 2 = 1 mol

=> \(m_{KNO_3}\) = 1. 101 = 101 g

b) Đổi 250 ml = 0,25l

\(n_{CaCl_2}\) = 0,25. 0,1 = 0,025 mol

=> \(m_{CaCl_2}\) = 0,025 . 111 = 2,775g

a) \(C\%_{ddKCl}=\dfrac{20}{500}.100\%=4\%\)

b) \(m_{dd}=15+45=60\left(g\right)\)

\(C\%_{ddNaCl}=\dfrac{15}{60}.100\%=25\%\)

a)

C% KCl = 20/500  .100% = 4%

b)

m dd = m NaCl + m H2O = 15 + 45 = 60(gam)

C% NaCl = 15/60  .100% = 25%

a)C%KCl=20/500×100=4%

b)C%NaCl=15/(15+45)×100=25%

Bn ko bt lm thì cũng đừng dùng chat gtp 

a)

\(C\%_{dd_{HCl}}=\dfrac{m_{HCl}}{mdd_{HCl}}\cdot100\%=\dfrac{20}{600}\cdot100\%=3,33\%\)

b)

\(mdd_{NaCl}=m_{NaCl}+m_{H_2O}=15+45=60\left(g\right)\)

=> \(C\%dd_{NaCl}=\dfrac{m_{NaCl}}{mdd_{NaCl}}\cdot100\%=\dfrac{15}{60}\cdot100\%=25\%\)

hok tốt nhé

Bài 1:

CuO + H2SO4 ---> CuSO4 + H2O
0,2____0,2_______0,2

mCuSO4 = 0,2.160 = 32g
mH2SO4 = 0,2.98 = 19,6g

mdd H2SO4 bđầu = mH2SO4/20% = 98g
mdd sau p/ứ = 98 + 0,2.80 = 114
mH2O = 114 - 32 = 82g

Gọi x là số mol CuSO4.5H2O tách ra

Cứ 100g H2O hòa tan được 17,4g CuSO4
=> (82-5x.18)g H2O hòa tan được (32-160x)g CuSO4
=> 100.(32-160x) = 17,4(82-5x.18) => x = 0,123mol

Vậy khối lượng CuSO5.5H2O tách ra là: 0,123.250 = 30,71g

Câu 2:

a) nNaOH=20/40=0,5(mol)

VH2O=mdd/D=250/1=250(ml)=0,25(l)

=>CM=0,5/0,25=2(M)

b) nHCl = 26,88/22,4=1,2 (mol)
=>CM = 1,2/0,5=2,4(M)

c)nNa2CO3=n Na2CO3.10H2O = 28,6/286=0,1(mol)
=>CM= 0,1/0,2=0,5(M)