Đề thi vào 10 môn Hóa có đáp án (Mới nhất ) (Đề số 7)

$C{H}_4\text{ +\hspace{0.33em}}2{\text{O}}_2\stackrel{t^o}{\to }C{O}_2{\text{\hspace{0.33em}+ 2H}}_{2}O$

${\text{CaCO}}_3\text{\hspace{0.33em}+\hspace{0.33em}}2C{H}_{3}COOH\to \text{\hspace{0.33em} (}C{H}_{3}COO{)}_2{\text{Ca\hspace{0.33em}+ CO}}_2\uparrow {\text{\hspace{0.33em}+\hspace{0.33em}H}}_{2}O$

$〈\begin{array}{l}Ban đầu{\text{\hspace{0.33em}: CO}}_2+2NaOH\to N{a}_{2}C{O}_3+H_{2}O\\ Sau đó{\text{\hspace{0.33em}:\hspace{0.33em}CO}}_2{\text{\hspace{0.33em}+ Na}}_2{\text{CO}}_3{\text{\hspace{0.33em}+ H}}_2\text{O }\to 2{\text{NaHCO}}_3\end{array}$

$\begin{array}{l}1.{\text{\hspace{0.33em}Khí H}}_2{\text{\hspace{0.33em}: Zn}}_{\text{(răn)\hspace{0.33em}}}+H_{2}S{O_4}_{\left(dd\right)}\to ZnS{O}_{\text{4}}+H_2\uparrow \\ 2.KhíC{O}_2{\text{\hspace{0.33em}: CaCO}}_{3\text{\hspace{0.33em}(rắn)}}+2HC{l}_{\left(dd\right)}\to CaC{l}_2{\text{\hspace{0.33em}+ CO}}_2\uparrow +H_{2}O\\ 3.\text{\hspace{0.33em}Khí }{O}_2:2H_2{O}_{2\text{\hspace{0.33em}(dd)}}\stackrel{Mn{O}_2\text{\hspace{0.33em}rắnn\hspace{0.33em}}}{\to }2H_{2}O+O_2\uparrow \end{array}$

$\begin{array}{l}•\text{\hspace{0.33em} ý kiÕn cña b¹n lµ sai v× :}\\ {\text{1.\hspace{0.33em}NaCl\hspace{0.33em}r¾n\hspace{0.33em}kh«ng ph¶n øng vì dung dịch H}}_{2}S{O}_4\text{\hspace{0.33em}lõang v× : }\\ {\text{ \hspace{0.33em} NaCl}}_{\left(\text{r¾n}\right)}{\text{\hspace{0.33em}+ H}}_{2}S{O}_{4\text{\hspace{0.33em}(®Æc)}}\stackrel{t^o}{\to }HCl\uparrow {\text{\hspace{0.33em}+ NaHSO}}_4\\ \text{2.\hspace{0.33em}}Dï\text{\hspace{0.33em}cã khÝ HCl sinh ra ®i n÷a th× không thể thu khí \hspace{0.33em}HCl bằng ph­ương pháp đẩy n­ước như­ trong h×nh vÏ v×}\\ {\text{HCl tan tèt trong n­íc (Ph­ương pháp đẩy n­ước chỉ thu các khí ít tan trong nước) : HCl}}_{\left(k\right)}\stackrel{H_{2}O}{\to }{\text{\hspace{0.33em}HCl}}_{\left(dd\right)}\end{array}$

$\begin{array}{l}•\text{\hspace{0.33em} ý kiÕn cña b¹n lµ sai v× :}\\ {\text{1.\hspace{0.33em}NaCl\hspace{0.33em}r¾n\hspace{0.33em}kh«ng ph¶n øng vì dung dịch H}}_{2}S{O}_4\text{\hspace{0.33em}lõang v× : }\\ {\text{ \hspace{0.33em} NaCl}}_{\left(\text{r¾n}\right)}{\text{\hspace{0.33em}+ H}}_{2}S{O}_{4\text{\hspace{0.33em}(®Æc)}}\stackrel{t^o}{\to }HCl\uparrow {\text{\hspace{0.33em}+ NaHSO}}_4\\ \text{2.\hspace{0.33em}}Dï\text{\hspace{0.33em}cã khÝ HCl sinh ra ®i n÷a th× không thể thu khí \hspace{0.33em}HCl bằng ph­ương pháp đẩy n­ước như­ trong h×nh vÏ v×}\\ {\text{HCl tan tèt trong n­íc (Ph­ương pháp đẩy n­ước chỉ thu các khí ít tan trong nước) : HCl}}_{\left(k\right)}\stackrel{H_{2}O}{\to }{\text{\hspace{0.33em}HCl}}_{\left(dd\right)}\end{array}$

$\begin{array}{l}•Chia\text{\hspace{0.33em}®Ò}ul­îng\text{\hspace{0.33em}thÓ tÝch dung dÞch NaOH vµo 2 cèc thñy tinh chia ®é}\left\{\begin{array}{l}Cèc\text{\hspace{0.33em}1 : NaOH\hspace{0.33em}}(x\text{\hspace{0.33em}mol)}\\ Cèc\text{\hspace{0.33em}2 : NaOH (x\hspace{0.33em}mol)}\end{array}\right.\\ •{\text{\hspace{0.33em}Sục khí CO}}_2{\text{đên\hspace{0.33em}dư­ vào cốc 1 : CO}}_{2\text{\hspace{0.33em}(d­)}}+NaOH\to NaHC{O}_3\left(1\right)\\ \text{\hspace{0.17em} x }\to \text{ x }\\ •\text{\hspace{0.33em}Sau đó đô cốc 2 (chứa NaOH)\hspace{0.33em}}vào{\text{\hspace{0.33em}côc 1 : NaHCO}}_3+NaOH\to {\text{ Na}}_{2}C{O}_3+H_{2}O\text{\hspace{0.33em}(2)}\\ \text{\hspace{0.33em} x x }\\ \to {\text{\hspace{0.33em}Phản ứng (2) xảy ra vừa đủ nên không còn dư­ NaOH hoặc NaHCO}}_3\text{. }\end{array}$