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MicrosoftInternetExplorer4
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a)
Theo
đề bài, ta có :
mdd HCl = 100,8 x 1,19 =
119,952 (g)
Pthh : Zn +
2HCl --> ZnCl2 + H2
(mol) a 2a a a
ZnO +
2HCl --> ZnCl2 + H2O
(mol) b
2b b
Gọi a, b lần lượt là số mol Zn, ZnO trong hh.
Áp dụng ĐLBTKL :
mZnCl2 = mhh + mdd
HCl - mH2
= 65a + 81b + 119,952 – 2a
= 161,352 (g)=> 63a + 81b = 41,4 (1)Lại có : nHCl = 2a + 2b
= (119,952 x 36,5)/(100 x 36,5) =
1,19952 (mol) (2)Giải hệ phương trình (1), (2) ta được: a = 0,39892;
b = 0,20084
=> mhh = 65 x 0,39892 + 81 x 0,20084 = 42,19784 (g)
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MicrosoftInternetExplorer4
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b) Vì pư vừa đủ
nên dung dịch A chỉ có ZnCl2.
Theo các pthh => nZnCl2 = a + b = 0,59976
(mol)
=> C% dd A (ZnCl2) = (0,59976 x 136)/161,352 x
100% = 50,552%
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MicrosoftInternetExplorer4
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font-family:"Times New Roman";
mso-ansi-language:#0400;
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mso-bidi-language:#0400;}
a)
Theo
đề bài, ta có :
mdd HCl = 100,8 x 1,19 =
119,952 (g)
Pthh : Zn +
2HCl --> ZnCl2 + H2
(mol) a 2a a a
ZnO +
2HCl --> ZnCl2 + H2O
(mol) b
2b b
Gọi a, b lần lượt là số mol Zn, ZnO trong hh.
Áp dụng ĐLBTKL :
mZnCl2 = mhh + mdd
HCl - mH2
= 65a + 81b + 119,952 – 2a
= 161,352 (g)=> 63a + 81b = 41,4 (1)Lại có : nHCl = 2a + 2b
= (119,952 x 36,5)/(100 x 36,5) =
1,19952 (mol) (2)Giải hệ phương trình (1), (2) ta được: a = 0,39892;
b = 0,20084
=> mhh = 65 x 0,39892 + 81 x 0,20084 = 42,19784 (g)
Normal
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false
false
false
MicrosoftInternetExplorer4
/* Style Definitions */
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font-size:10.0pt;
font-family:"Times New Roman";
mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}
b) Vì pư vừa đủ
nên dung dịch A chỉ có ZnCl2.
Theo các pthh => nZnCl2 = a + b = 0,59976
(mol)
=> C% dd A (ZnCl2) = (0,59976 x 136)/161,352 x
100% = 50,552%
Normal
0
false
false
false
MicrosoftInternetExplorer4
/* Style Definitions */
table.MsoNormalTable
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font-size:10.0pt;
font-family:"Times New Roman";
mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}
a)
Theo
đề bài, ta có :
mdd HCl = 100,8 x 1,19 =
119,952 (g)
Pthh : Zn +
2HCl --> ZnCl2 + H2
(mol) a 2a a a
ZnO +
2HCl --> ZnCl2 + H2O
(mol) b
2b b
Gọi a, b lần lượt là số mol Zn, ZnO trong hh.
Áp dụng ĐLBTKL :
mZnCl2 = mhh + mdd
HCl - mH2
= 65a + 81b + 119,952 – 2a
= 161,352 (g)=> 63a + 81b = 41,4 (1)Lại có : nHCl = 2a + 2b
= (119,952 x 36,5)/(100 x 36,5) =
1,19952 (mol) (2)Giải hệ phương trình (1), (2) ta được: a = 0,39892;
b = 0,20084
=> mhh = 65 x 0,39892 + 81 x 0,20084 = 42,19784 (g)
Normal
0
false
false
false
MicrosoftInternetExplorer4
/* Style Definitions */
table.MsoNormalTable
{mso-style-name:"Table Normal";
mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
mso-style-noshow:yes;
mso-style-parent:"";
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mso-para-margin:0cm;
mso-para-margin-bottom:.0001pt;
mso-pagination:widow-orphan;
font-size:10.0pt;
font-family:"Times New Roman";
mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}
b) Vì pư vừa đủ
nên dung dịch A chỉ có ZnCl2.
Theo các pthh => nZnCl2 = a + b = 0,59976
(mol)
=> C% dd A (ZnCl2) = (0,59976 x 136)/161,352 x
100% = 50,552%