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\(a.2KClO_3\xrightarrow[t^0]{xt}2KCl+3O_2\\ b.n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{22,05}{122,5}=0,18mol\\ n_{O_2\left(lt\right)}=\dfrac{0,18.3}{2}=0,27mol\\ V_{O_2\left(lt\right)}=n_{O_2\left(lt\right)}.22,4=0,27.22,4=6,048l\\ H=\dfrac{V_{O_2\left(tt\right)}}{V_{O_2\left(lt\right)}}\cdot100\%=\dfrac{3,36}{6,048}\cdot100\%\approx55,56\%\\ c.n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\\ n_{KCl}=\dfrac{0,15.2}{3}=0,1mol\\ m_{KCl}=n_{KCl}.M_{KCl}=0,1.74,5=7,45g\)
\(a,PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ b,n_{KClO_3(thực tế)}=\dfrac{36,75}{122,5}=0,3(mol)\\ n_{O_2(phản ứng)}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{KClO_3(phản ứng)}=\dfrac{2}{3}n_{O_2}=0,2(mol)\\ \Rightarrow H\%=\dfrac{0,2}{0,3}.100\%=66,67\%\\ c,n_{KCl}=n_{KClO_3(phản ứng)}=0,2(mol)\\ \Rightarrow m_{KCl}=0,2.74,5=14,9(g)\)
a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)
c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(a,2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ b,n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\\ n_{MgO}=n_{Mg}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ b,x=m_{MgO}=40.0,4=16\left(g\right)\\ V=V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{KClO_3}=\dfrac{2}{15}.122,5=\dfrac{49}{3}\left(g\right)\)
a, 2Mg + O2 \(\rightarrow\) 2MgO (bạn thêm to trên cái mũi tên nhé)
b, nMg = \(\dfrac{9,6}{24}\) = 0,4 (mol)
PTPƯ: 2Mg + O2 \(\rightarrow\) 2MgO
2g/mol 1g/mol 2g/mol
\(\Rightarrow\) 0,4 0,2 0,4
VO2 = 0,2 . 22,4 = 4,48l
mMgO = 0,4 . (24 + 16) = 16(g)
2)
nH2 = 6.72/22.4 = 0.3 (mol)
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
0.1______0.3______0.2
mFe2O3 = 0.1*160 = 16 (g)
mFe = 0.2*56 = 11.2 (g)
3)
nFe3O4 = 11.6/232 = 0.05 (mol)
3Fe + 2O2 -to-> Fe3O4
0.15___0.1______0.05
mFe = 0.15*56 = 8.4 (g)
VO2 = 0.1*22.4 = 2.24 (l)
2KClO3 -to-> 2KCl + 3O2
1/15______________0.1
mKClO3 = 1/15 * 122.5 = 8.167 (g)
a)
3H2 + Fe2O3 --to--> 2Fe + 3H2O
b) nH2 = 6,72/22,4 = 0,3 mol
Từ pt => nFe3O4 = 0,1 mol
=> mFe3O4 = 0,1. 232 = 23,2 g
1. a. \(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\left(1\right)\)
b. Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)
Theo PT(1): \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,1=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(lít\right)\)
c. \(PTHH:2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\left(2\right)\)
Theo PT(2): \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=40,83\left(g\right)\)
2. \(PTHH:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
a. Theo PT: \(n_{Fe}=3.n_{Fe_3O_4}=0,01.3=0,03\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)
b. Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,02.22,4=0,448\left(lít\right)\)
a)PTHH:\(2KCLO_3\rightarrow^{t^o}2KCL+3O_2\)
b)\(n_{KCL}=\frac{2,98}{74,5}=0,4mol\)
Từ phương trình hóa học\(n_{KCL}=n_{KCLO_3}=0,04mol\)
\(n_{KCLO_3}=0,04\cdot122,5=4,9g\)
\(H=\frac{4,9}{9,8}\cdot100\%=50\%\)
a. PTHH: \(2KClO_3\rightarrow^{t^o}2KCl+3O_2\)
b. \(n_{KCl}=\frac{2,98}{74,5}=0,4mol\)
Từ phương trình hoá học \(n_{KCl}=n_{KClO_3}=0,04mol\)
\(n_{KClO_3}=0,04.122,5=4,9g\)
\(H=\frac{4,9}{9,8}.100\%=50\%\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
Công thức khối lượng :
\(m_P+m_{O_2}=m_{P_2O_5}\)
Khi đó :
\(m_P=m_{P_2O_5}-m_P=14.2-6.2=8\left(g\right)\)
a) \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)
b) \(m_P+m_{O_2}=m_{P_2O_5}\)
c) áp dụng định luật bảo toàn khối lượng, ta có:
\(m_P+m_{O_2}=m_{P_2O_5}\)
\(\Rightarrow m_{O_2}=m_{P_2O_5}-m_P=14,2-6,2=8\left(g\right)\)
vậy khối lượng oxi đã phản ứng là \(8g\)
\(2KClO_3\rightarrow3O_2+2KCl\)
\(m_{KClO_3}=m_{O_2}+m_{KCl}\)
\(\Rightarrow m_{KCl}=m_{KClO_3}-m_{KCl}=24,5-9,6=14,9\left(g\right)\)
mik làm ròi nhé
cho mk xin đc ko ạ
cảm ơn nhìu nha