tinh khoi luong ruou etylic can dung de dieu che 200g dd ch3cooh tren bang phuong phap len men voi hieu suat 80%
cho x gam glucozo len men thi thu duoc ruou etylic va mot chat khac,dau khi sinh ra vao nuoc voi trong thu duoc thu duoc 80g ket tua a,tim x,biet hieu suat phan ung la 80% b,tinh khoi luong ruoi thu duoc
bạn ghi đề có vài chỗ mik ko hỉu lắm "dau khi sinh ra vào nước vôi trong..."???
Cau 1: hoa tan 4g NaOH vao 200ml nuoc. Tinh nong do mol/l cua dd thu duoc. Tinh nong do % cua dd thu dc. Tinh do pH cua dd thu dc. Cho 100ml dd thu dc o tren tac dung voi 100ml dd CuSO4 1M. Tinh k.luong ket tua thu dc
Cau2: dot chay 600g 1 mau than da chua tap chat khong chay thu dc 1met khoi khi CO2(dktc). Tinh % ve k.luong cacbon trong than
Cau3: de dieu che 500g TNT can dung bao nhieu gam toluen biet hieu suat phan ung la 90%
Cau4: cho 6g axit axetit tac dung 1 luong du ancol etilit co H2SO4 dac lm xuc tac. Tinh k.luong este thu dc biet hieu suat la 65%
Cho 300g dung dich CH3COOH tac dung vua du voi 500g dung dich NaHCO3 20 phan tram.
a,Hay tinh nong do phan tram cua dung dich CH3COOH.
b,Hay tinh the tich cua khi butan can de dieu che duoc luong CH3COOHco trong 300g dung dich tren.
tinh khoi luong AlO2 can dung de san xuat 2,7 tan nhom voi hieu xuat khoang 80%
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ BTNT\left(Al\right):n_{Al_2O_3}.2=n_{Al}\\ \Rightarrow n_{Al_2O_3}=0,05\left(mol\right)\\VìH=80\%\\ \Rightarrow m_{Al_2O_3}=\dfrac{0,05.102}{80\%}=6,375\left(tấn\right)\)
Len men a(g) glucozo voi hieu suat 80%. Cho toan bo luong khi CO2 sinh ra hap thu het vao dung dich CA(OH)2 thu duoc 10g ket tua. Khoi luong dung dich sau phan ung giam 3,4g so voi ban dau. Tim a.
LAM ON GIAI NHANH GIUP MINH BAI NAY VOI!!!!! CAN GAP NHE!!!! CAM ON NHIEU!!!!
de dieu che khi oxi nguoi ta da dung kclo3 nhiet phan
a, viet phuong trinh phan ung tren
b, tinh the tich khi oxi thu duoc ( o dieu kien tieu chuan )khi nhiet phan 73, 5 g kclo3
c tinh khoi luong zno duoc tao thanh khi cho luong khi cho luong khi oxi sinh ra o tren tac dung vs 13 g zn
giai de hoa hoc trong phong thi nghiem oxi duoc dieu che bang cach nhiet KCLO3 hoac KMNO4 Tinh khoi luong KCLO3 Hoac KMNO4 Can dung de dieu che duoc 0.72 lit khi oxi o DKTC (cho O=16, Cl = 35,5 , K = 39 , Mn = 55)
\(n_{O_2} = \dfrac{0,72}{22,4} = \dfrac{9}{280}(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = \dfrac{9}{140}(mol)\\ m_{KMnO_4} = \dfrac{9}{140}.158 = 10,16(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{3}{140}(mol)\\ m_{KClO_3} = \dfrac{3}{140}.122,5 = 2,625(gam)\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.6..............................................0.3\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(0.2..........................0.3\)
\(m_{KMnO_4}=0.6\cdot158=94.8\left(g\right)\)
\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)
\(n_{O_2}=\dfrac{V_{O_2\left(ĐKTC\right)}}{22,4}=\dfrac{0,72}{22,4}=0,03\left(mol\right)\)
- Điều chế O2 bằng KMnO4, ta có PTHH:
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
----2----------------------------------------------------1---
----0,06------------------------------------------------0,03---
\(m_{KMnO_4}=n_{KMnO_4}\cdot M_{KMnO_4}=0,06\cdot158=9,48\left(g\right)\)
- Điều chế O2 bằng KClO3, ta có PTHH:
\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)
---2-------------------------------------3----
---0,02---------------------------------0,03----
\(m_{KClO_3}=n_{KClO_3}\cdot M_{KClO_3}=0,02\cdot122,5=2,45\left(g\right)\)
TIH KHOI LUONG BENZEN CAN DUNG DE DIEU CHE 39,25 G BROMBENZEN .BIET HIEU XUAT PHAN UNG DAT 85%
Tính khối lượng axit axetic va ancol etylic can dung de dieu che duoc 6.6 g etyl axetat biet hieu suat phan ung este la 75%
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$n_{CH_3COOC_2H_5} = \dfrac{6,6}{88} = 0,075(mol)$
$n_{CH_3COOH\ đã\ dùng} = n_{C_2H_5OH\ đã\ dùng} = 0,075 :75\% = 0,1(mol)$
$m_{CH_3COOH} = 0,1.60 = 6(gam)$
$m_{C_2H_5OH} = 0,1.46 =4 ,6(gam)$