\(Zn+2HCl->ZnCl_2+H_2\\ m_{Zn}=\dfrac{7,437}{24,79}\cdot65=19,5g\\ m_{HCl}=\dfrac{7,437}{24,79}\cdot2\cdot36,5=21,9g\)

\(n_{glu}=a;n_{suc}=b\\ n_{O_2}=6a+12b=0,48\\ 2a+4b=0,16\\ glucose->2Ag\\ sucrose->glucose+fructose->4Ag\\ m_1=108\left(2a+4b\right)=108\cdot0,16=17,28\left(g\right)\)

\(\left(C_6H_{10}O_5\right)_n->nC_6H_{12}O_6->2nCO_2\\\sum n_{CO_2}=\dfrac{10}{100}+2\cdot\dfrac{5}{100}=0,2mol\\ m=\dfrac{0,2\cdot162n}{2n\cdot0,9\cdot0,8}=22,5\left(g\right)\)

\(Zn+CuSO_4->ZnSO_4+Cu\\ \Delta m\downarrow=0,12\cdot6,1\left(65-64\right)=0,732g\)

\(Na_2O+H_2O->2NaOH\\ n_{NaOH}=2\cdot\dfrac{5,6}{52}=0,18mol\\ C_M=\dfrac{0,18}{3}=0,06M\\ 0,06=\dfrac{10\cdot1,2\cdot C\%}{40}\\ C\%=20\%\)

bạn ơi đc r sao kêu bồ ơi=))

A: `N_2`

C: `NO`

D: `NO_2`

E: `HNO_3`

G: `NaNO_3`

\(\left(1\right)2NH_3+3CuO\underrightarrow{t^o}3Cu+N_2+3H_2O\)

\(\left(2\right)N_2+3H_2\underrightarrow{xt,t^o}2NH_3\)

\(\left(3\right)4NH_3+5O_2\underrightarrow{xt,t^o}6H_2O+4NO\)

\(\left(4\right)2NO+O_2\rightarrow2NO_2\)

\(\left(5\right)4NO_2+O_2+2H_2O\rightarrow4HNO_3\)

\(\left(6\right)HNO_3+NaOH\rightarrow NaNO_3+H_2O\)

19

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ n_{H_2SO_4}=0,06.1=0,06\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,1------->0,05------->0,05

Xét \(\dfrac{0,1}{2}< \dfrac{0,06}{1}\Rightarrow H_2SO_4\) dư

a. A gồm \(\left\{{}\begin{matrix}n_{H_2SO_4dư}=0,06-0,05=0,01\left(mol\right)\\n_{Na_2SO_4}=0,05\left(mol\right)\end{matrix}\right.\)

Trong A: \(\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,01}{0,06}=\dfrac{1}{6}M\\CM_{Na_2SO_4}=\dfrac{0,05}{0,06}=\dfrac{5}{6}M\end{matrix}\right.\)

b. Nhúng giấy quỳ tím vào dung dịch (A) quỳ chuyển sang màu đỏ vì axit `H_2SO_4` dư

20

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1--------------->0,2

a.

\(a=CM_{NaOH}=\dfrac{0,2}{0,2}=1M\)

b.

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------->0,1

\(V_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)