Hóa học Lớp 11: Đốt hoàn toàn a gam photpho trong oxi dư, rồi cho sản phẩm tác dụng với dung dịch NaOH dư, thu được 82,0 gam muối. Giá trị của a là 800×600 Normal 0 false false false EN-US X-NONE X-NONE 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-priority:99; mso-style-parent:””; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:”Calibri”,”sans-serif”; mso-bidi-font-family:”Times New Roman”;} A. 82,0. 800×600 Normal 0 false false false EN-US X-NONE X-NONE 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-priority:99; mso-style-parent:””; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:”Calibri”,”sans-serif”; mso-bidi-font-family:”Times New Roman”;} B. 31,0 800×600 Normal 0 false false false EN-US X-NONE X-NONE 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-priority:99; mso-style-parent:””; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:”Calibri”,”sans-serif”; mso-bidi-font-family:”Times New Roman”;} C. 15,5 800×600 Normal 0 false false false EN-US X-NONE X-NONE 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-priority:99; mso-style-parent:””; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:”Calibri”,”sans-serif”; mso-bidi-font-family:”Times New Roman”;} D. 46,5, giúp em giải bài này ạ, em cảm ơn thầy cô và các bạn nhiều.
Giải đáp C
4P + 5O2 → 2P2O5
P2O5 + 6NaOH dư → 2Na3PO4 + 3H2O
=> nNa3PO4 = 82/164 = 0,5 (mol)
BTNT P => nP = nNa3PO4 = 0,5 (mol) => mP = 0,5.31 = 15,5 (gam)