112al +6hcl = 2alcl3 +3 h2
Al. At.wt = 27
2* 27 = 3*2
5 = x
x =5*6/54
x = 5/9
x = 5/9
2g is 22.4l(dm3)
5/9 is 6.22 dm3
Therefore the answer is 6.22dm3
Editted after finding my mistake (stupid foolish mistake)
5g of alumunium treated with excess of dilute HCl ,the volume of
hydrogen evolved at STP in dm3
Ans is 6.22......can u tell me how
euql masses of zinc n bromine were allowed to react till
completion of reaction to form ZnBr2 which substance is left
unreacted n to wat % of its original mass
ans-zinc-59.3 urgent plz,,,,,,,,,,,,,,,,
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10 Answers
11
13572Al+6HCl→2AlCl3+3H2
mole of Al taken=5/27
moles of hydrogen formed=(5/27)*3/2=5/18
volume of hydrogen=(5/18)*22.4=6.22 dm3
112.
zn + br2 = znbr2
Therefore
65g of zinc will react with 160g of Br
Therefore we take 100g each
160 65
100 ?
100*65/160
x = 40.625g will be raeacted Therefor the remaining mass is 100- 40.625 = 59.375g
Therefore the percentage of the remainder is 59.375%
1under which condition the density of a gas is minimum
STP
0 degree C under 2 atm
100 degree c under 2 atm
100 degree c 0.5 atxplain
1357PV=nRT
P=(n/V)RT=(m/MV)RT=dRT/M
d=MP/RT
so d=KP/T
now try to find when it will be maximum
11D = M/V
DV is a constant ratio
Therefore if volume increases D reduces
Now
PV = nRT
nR is constant assume K
V = KT/P
Assume STP volume to be V0
Now in 2) Temperature is same as STP but pessure is increased therefore V1<V0
V 1 = K * 273 /2
3) V2>V1
V 2 = K * 373/2
Now check the no. 4.
V4 = K * 373 * 2
V0 = K * 273
Therefore V4>V0>V2>V1
Therefore The density will be minimum for 100°C and 0.5 atm
1on treatment with dilute alkali two molecules of acetone combine to form
acetone alcohol
diacetone
diactone alcohol
acetal
