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\(a,m_{dd}=\dfrac{5}{12\%}=\dfrac{125}{3}\left(g\right)\\ b,m_{dd}=\dfrac{4}{7,3\%}=\dfrac{4000}{73}\left(g\right)\\ c,m_{NaOH}=0,5.40=20\left(g\right)\\ m_{dd}=\dfrac{20}{10\%}=200\left(g\right)\)
\(a,C\%_{NaCl}=\dfrac{15}{15+185}.100\%=7,5\%\\ b,m_{HNO_3}=\dfrac{18,9}{100}.100+\dfrac{6,3}{100}.200=31,5\left(g\right)\\ m_{ddHNO_3}=100+200=300\left(g\right)\\ C\%_{HNO_3}=\dfrac{31,5}{300}.100\%=10,5\%\)
\(c,n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\\ C_{M\left(NaCl\right)}=\dfrac{0,1}{0,1}=1M\\ d,n_{KOH}=2.0,2+0,2.0,2=0,44\left(mol\right)\\ V_{ddKOH}=0,2+0,2=0,4\left(l\right)\\ C_{M\left(KOH\right)}=\dfrac{0,44}{0,4}=1,1M\\ e,m_{NaOH}=\dfrac{150.16}{100}=24\left(g\right)\\ m_{ddNaOH}=50+150=200\left(g\right)\\ C\%_{NaOH}=\dfrac{24}{200}.100\%=12\%\)
\(a,m_{NaCl}=\dfrac{8.200}{100}=16\left(g\right)\\ b,m_{HCl}=\dfrac{250.14}{100}=35\left(g\right)\\ m_{H_2SO_4}=\dfrac{19,6.300}{100}=58,8\left(g\right)\)
\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ n_{H_2SO_4}=3.0,15=0,45\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,45.98.100}{14,7}=300\left(g\right)\\ m_{ddsau}=24+300=324\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\\ C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,15.400}{324}.100\approx18,519\%\)
\(a,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,2--------------->0,4
b, \(V_{ddKOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, dd sau phản ứng là QT chuyển sang màu xanh vì KOH là dd bazơ
\(n_{K_2O}=\dfrac{11,28}{94}=0,12\left(mol\right)\)
PTHH: K2O + H2O → 2KOH
Mol: 0,12 0,24
mKOH=0,24.56 = 13,44 (g)
mddA = 200+11,28 = 211,28 (g)
\(\Rightarrow C\%_{ddA}=\dfrac{13,44.100\%}{211,28}=6,36\%\)
a)
\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)
=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)
=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)
b)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)
=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)
=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)
1
\(a)m_{H_2O}=250-5=245g\\b )C_{\%NaCl}=\dfrac{5}{250}\cdot100=2\%\)
\(2\\ m_{ddCuSO_4}=\dfrac{15.100}{5}=300g\\ m_{H_2O}=300-15=285g\)
Câu 1:
a, Ta có: m dd = m chất tan + mH2O ⇒ mH2O = 250 - 5 = 245 (g)
b, \(C\%_{NaCl}=\dfrac{5}{250}.100\%=2\%\)
Câu 2:
Ta có: \(C\%_{CuSO_4}=\dfrac{15}{m_{ddCuSO_4}}.100\%=5\%\)
\(\Rightarrow m_{ddCuSO_4}=300\left(g\right)\)
⇒ mH2O = 300 - 15 = 285 (g)
Câu 3:
Theo đề bài ta có :
nZn=\(\dfrac{9,75}{65}=0,15\left(mol\right)\)
\(mHCl=\dfrac{m\text{dd}.C\%}{100\%}=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)
=> nHCl=\(\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Ta có pthh
-----Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
0,15mol...0,3mol....0,15mol..0,15mol
Theo pthh ta có tỉ lệ :
nZn=\(\dfrac{0,15}{1}mol< nHCl=\dfrac{0,4}{2}mol\)
=> số mol của HCl dư ( tính theo số mol của Zn)
Dung dịch thu được sau phản ứng bao gồm ddHCl(dư) và ddZnCl2
=> mct=mZnCl2=0,15.136=20,4 g
mHCl(dư)=(0,4-0,3).36,5=3,65 g
mddZnCl2=mZn + mddHCl - mH2 = 9,75+200-(0,15.2)=209,45 g
=> C%\(_{ZnCl2}=\dfrac{mct}{m\text{dd}}.100\%=\dfrac{20,4}{209,45}.100\%\approx9,74\%\)
C%\(_{HCl\left(d\text{ư}\right)}=\dfrac{mct}{m\text{dd}}.100\%=\dfrac{3,65}{209,45}.100\%\approx1,743\%\)
Câu 1:
Theo đề bài ta có:
nK2O=\(\dfrac{12,4}{94}\approx0,13\left(mol\right)\)
mKOH(bđ)=\(\dfrac{m\text{dd}.C\%}{100\%}=\dfrac{120.5\%}{100\%}=6\left(g\right)\)
=> nKOH(bđ)=\(\dfrac{6}{56}\approx0,107\left(mol\right)\)
ta có pthh
K2O + H2O \(\rightarrow\) 2KOH
0,13mol...........0,26mol
=> mdd(thu-được)=12,4+120=132,4(g)
Dung dịch thu được sau phản ứng là KOH
Ta có
nKOH(thu-được)= 0,26+0,107 = 0,367 mol
=> mct=mKOH(thu-được)=0,367.56=20,552(g)
=> C%\(_{\left(dung-d\text{ịch}-thu-\text{đ}\text{ư}\text{ợc}\right)}=\dfrac{mct}{m\text{dd}}.100\%=\dfrac{20,552}{132,4}.100\%\approx15,523\%\)