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Nostrana [21]
3 years ago
10

How many liters of .100 m hcl hcl would be required to react with 5 grams of calcium hydroxide?

Chemistry
2 answers:
aniked [119]3 years ago
7 0
The balanced equation for the reaction between HCl and Ca(OH)₂ is as follows;
Ca(OH)₂ + 2HCl ---> CaCl₂  + 2H₂O
stoichiometry of Ca(OH)₂ to HCl is 1:2
mass of Ca(OH)₂ reacting - 5 g
therefore number of moles of Ca(OH)₂ - 5 g / 74 g/mol = 0.068 mol 
according to molar ratio
number of HCl moles reacted = twice the number of Ca(OH)₂ reacted 
number of HCl moles required - 0.068 x 2 = 0.136 mol 
molarity if HCl solution - 0.100 M
there are 0.100 mol of HCl in 1 L 
therefore 0.136 mol in - 0.136 mol / 0.100 mol/L = 1.36 L
volume of 0.100 M HCl required - 1.36 L 
valkas [14]3 years ago
7 0
The  number  of liters  of  0.100 HCl that would  be required  to  react with  5g of Ca(OH)2 is calculated  as  follows

find the   moles of Ca(OH)2  used

= mass/molar mas
=  5g/ 74.09 g/mol  = 0.0675  moles
write the  equation  involved

that is Ca(OH)2 +2 HCl = CaCl2 + 2 H2O

by use of mole ratio  between  Ca(OH)2  to HCl  which  is 1:2  the  moles  of  HCl  is therefore = 2 x 0.0675 = 0.135  moles  of HCl

volume of HCl =  moles of HCL / molarity of HCl

= 0.135/ 0.100 = 1.35 L
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A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
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T = temperature of gas = 270^oC=273+270=543K

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Now put all the given values in the ideal gas equation, we get:

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Now we have to calculate the volume at 24°C.

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where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

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where,

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Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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