1) Tron 150 ml dd NaOH 1M voi 100 ml dd KOH 0,5M thu duoc dung dich C
a) Tinh nong do cac ion trong dd C
b) Trung hoa dd C bang 200 ml dd H2SO4 co nong do mol la a mol/ lit. Tinh a
1) Tron 100 ml dd HCl 1M voi 100 ml dd H2SO4 0,5M thu duoc dd D
a) Tinh nong do cac ion trong dd D
b) Cho dd D tac dung voi dd BaCl2 du thu duoc m gam ket tua. Tinh m
\(a.n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ \left[HCl\right]=\dfrac{0,1}{0,1+0,1}=0,5\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,05}{0,1+0,1}=0,25\left(M\right)\\ \left[H^+\right]=0,5+0,25.2=1\left(M\right)\\ \left[SO^{2-}_4\right]=\left[H_2SO_4\right]=0,25\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=0,5\left(M\right)\)
\(b.BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{\downarrow}=m_{BaSO_4}=233.0,05=11,65\left(g\right)\)
tron 100ml dd h2so4 co nong do x(M) voi 150ml dd NaOH 0,2 M thu duoc 200 ml dung dich A. Biet 100ml dd A hoa tan vua du 0,204 mol Al2O3. Tinh x
Trung hoa hoan toan v ml dd HNO3 1M can dung vua du 200ml dd NAOH 0,5M thu duoc dd X. Xac dinh gia tri cua V va tinh nong do mol cua cac ion trong dung dich X
\(n_{H^+}=n_{HNO_3}=V\)mol
\(n_{OH^-}=n_{NaOH}=0,5.0,2=0,1\) mol
\(H^++OH^-\rightarrow H_2O\)
0,1<--0,1
\(\Rightarrow n_{H^+}=V=0,1\)lít = 100 ml
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
0,1 -----> 0,1 ---------->0,1
\(NaNO_3\rightarrow Na^++NO_3^-\)
\(\Rightarrow\left[Na^+\right]=\left[NO_3^-\right]=\dfrac{0,1}{0,1+0,2}=0,33M\)
tinh nong do mol/lit cua cac ion
tron lan 15 ml dd naoh 2m voi 10ml dd h2so4 3m
1) a) Hoa tan 12,5 gam tinh the CuSO4. 5H2O trong nuoc thanh 200ml dd. Tinh nong do mol cac ion trong dd thu duoc
b) Hoa tan 8,08 gam Fe(NO3)3.9H2O trong nuoc thanh 500 ml dd. Tinh nong do mol cac ion trong dd thu duoc
a)cho 1 dd a co chua 98g h2so4 trong 1l dd phai them bao nhieu lit dd a vao 0,4 lit dd h2so4 nong do 2m de duoc dung dich x biet rang 100 ml dd x khi tac dung voi dd ba(oh)2 du cho ra 32.62 ket tu
b)lay 200 g dung dich x noi tren cho td voi dd y chua 2 bazo naoh 0.5m va koh 0.9m phai dung bao nhieu lit dd y de phan ung vu du voi 200ml dd x phan ung giua 2dd x ,y cho ra dd z tinh khoi luong muoi khan thu duoc sau khi co can dd
c) tron 0.4 dd x voi 1l dd y noi tren douc dd q co tinh axit hay bazo co the td toi da voi bao nhieu gam Al va fe (chi 1 trong 2 kim loai nay phan ung voi Q) tinh the tich khi bay ra dktc
\(n_{BaSO_4}=\frac{m}{M}=\frac{32,62}{233}=0,14mol\)
PTHH:
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)
0,14 0,14 0,14 0,28 (mol)
Gọi \(V_{ddH_2SO_4}\)cần thêm là x
\(n_{H_2SO_4}=\frac{m}{M}=\frac{98}{98}=1mol\)
\(C^{\left(A\right)}_{M_{H_2SO_4}}=\frac{1}{1}=1M\)
\(n^{\left(A\right)}_{H2SO4}=C_M.V=1.x=xmol\)
\(n_{H2SO4}=C_M.V=2.0,4=0,8mol\)
\(C_{MX}=\frac{n}{V}=\frac{0,8+x}{0,4+x}\left(M\right)\)
\(n_X=C_{MX}.V\)
\(\Leftrightarrow0,14=\frac{0,8+x}{0,4+x}.0,1\)
\(\Leftrightarrow\frac{0,14}{0,1}=\frac{0,8+x}{0,4+x}\)
⇔0,08+0,1x=0,56+0,14x
⇔x=0,6(l)
Vậy cần thêm 0,6 l dung dịch
DE CUONG ON TAP HK1 MON HOA HOC
1. hoa tan 0,54 gam nhom vao dd 120 gam dd h2so4 4,9% thoat ra V lit khi hidro (dktc)
a) viet pthh . tinh V ?
b) tinh nong do phan tram cua cac chat trong dd sau phan ung ?
2. hoa tan 15,5 gam na2o vao nuoc thanh 500 ml dung dich A
a) viet pthh xay ra
b) tinh nong do mol cua dd A
c) tinh the tich dd h2so4 20% ( D = 1,14g/ml ) can de trung hoa luong dd tren
1.
Theo đề bài ta có : \(\left\{{}\begin{matrix}nAl=\dfrac{0,54}{27}=0,02\left(mol\right)\\nH2SO4=\dfrac{120.4,9}{100.98}=0,06\left(mol\right)\end{matrix}\right.\)
PTHH :
\(2Al+3H2SO4->Al2\left(So4\right)3+3H2\uparrow\)
0,02mol...0,03mol.......0,01mol.............0,03mol
Theo PTHH ta có : nAl = \(\dfrac{0,02}{2}mol< nH2SO4=\dfrac{0,06}{2}mol=>nH2SO4\left(dư\right)\) ( tính theo nal)
=> VH2(đktc) = 0,03.22,4 = 6,72(l)
=> \(\left\{{}\begin{matrix}C\%ddH2SO4\left(dư\right)=\dfrac{\left(0,06-0,03\right).98}{0,54+120-0,03.2}.100\%\approx2,44\%\\C\%ddAl2\left(SO4\right)3=\dfrac{0,01.302}{0,54+120-0,03.2}.100\%\approx2,5\%\end{matrix}\right.\)
Theo đề bài ta có : nNa2O = \(\dfrac{15,5}{62}=0,25\left(mol\right)\)
a) PTHH :
\(Na2O+H2O->2NaOH\)
0,25mol....0,25mol.....0,5mol
b) Nồng độ mol dd A là :
CMddNaOH = 0,5/0,5 = 1(M)
c) PTHH :
\(2NaOH+H2SO4->Na2SO4+H2O\)
0,5mol.........0,25mol
=> mddH2SO4 = \(\dfrac{0,25.98}{20}.100=122,5\left(g\right)=>VddH2SO4=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)
1.Cho 1,68 lit CO2 (dktc) suc vao binh dung dd KOH du. Tinh nong do mol/l cua muoi thu dc sau phan ung. Biet rang the tich dd la 250 ml.
2.Cho 11, lit CO2 (dktc) vao dd NaOH 25% (d=1,3g/ml). Tinh nong do mol/l cua dd muoi tao thanh.
3.Dan 448ml CO2 (dktc) suc vao binh chua 100ml dd KOH 0,25M. Tinh khoi luong muoi tao thanh.
4.Hap thu 2.24l SO2 (dktc) vao 1 lit dung dich Ca(OH)2 0,06M. Tinh khoi luong muoi thu dc.
( Khong co dau hoi kho nhin, xl nha. Mk can gap trong ngay hom nay luon, cac bn giup mk nha, cam on nhieu !!!)
1.
nCO2=0,075mol
do dư KOH nên tạo ra muối trung hòa
nNa2CO3=nCO2=0,075mol
-->Cm=0,3M
2.
nCO2=0,5mol, nNaOH=4,0625mol
-->tạo muối Na2CO3 với số mol =nCO2=0,5mol
-->CmNa2CO3=1M
3.nCO2=0,02mol, nKOH=0,025mol
-->T=1,25 hay 1<T<2
tạo 2 muối
K2CO3:x, KHCO3:y
x+y=0,02
2x+y=0,025
-->x=0,005mol, y=0,015mol
-->mK2CO3=0,69g
mKHCO3=1,5g
Tron 200ml dung dich Ba(OH)2 0,05M voi 300ml dung dich HCl co pH=1, sau phan ung thu duoc A:
a. Tinh nong do mol/l cac ion trong dd A? tinh pH dd A?
b. them 250ml dd gom H2SO4 0,1M va Na2SO4 0,1M vao dd A sau phan ung thu duoc dd B va ket tua C . Tinh khoi luong ket tua va so mol cac ion trong dd B?
\(a)n_{Ba\left(OH\right)_2}=0,05\cdot0,2\cdot2=0,02mol\\ pH=1\Rightarrow\left[OH^-\right]=0,1M\Rightarrow n_{HCl}=0,1\cdot0,3=0,03mol\\ n_{Ba\left(OH\right)_2}+n_{HCl}=0,02+0,03=0,05mol\\ \Rightarrow C_M=\dfrac{0,05}{0,5}=0,1M\Rightarrow pH=1\)