How much of magnesium is required to react with 260cm3 of 0.5 M HCl? [Mg = 24]

Oluwadecent

7 years ago

2 moles of mg = 2x24= 48
conc. in mole of Hcl x molar mass x 1000(per dm3)/ mass of mg x 260
therefore 0.5x36.5x1000/260x48
=1.46g
approx=1.5g

galaxyxy

6 years ago

Mg(s) + 2HCL(aq) --> MgCl2(aq)+ H2(g)
1mole of Mg react with 2moles of HCL
So to find reactive moles of HCL - Molar conc.= Amount/Volume
Volume =250cm³ =0.25dm³
Molar conc.=0.5M
Therefore Amount = Molar conc. * Volume
Amount = 0.5 *0.25=0.125mols
Then to find reactive mol of Mg
Knowing that 1 mol react with 2 mols of HCL
0.125mol of HCL/2 = 0.0625mol of Mg
Therefore 0.625mol of Mg will react with 0.125mols of HCL
To find the required mass of Mg
Using Amount = Mass/Molar mass
Mass = Amount*Molar mass
Mass= 0.0625*24
Mass=1.5g

Alisha101

2 years ago

Firstly make a equation about the given question
Mg+2HCl=MgCl2+H2
then No of moes= Concentration in mol/dm3*volume
0.25*0.5= 0.125
1:2
x:0.125
2x=0.125
x=0.0625
Recall;no of moles = reacting mass / molar mass
reacting mass= 0.0625*24
m=1.5g

De sage

1 year ago

first and foremost
mole= CV/1000
mole= 0.5×260/1000
mole= 0.13m

Mg+ 2HCl ------>MgCl2+H2

24g of Mg ----> 2moles of HCl
xg of Mg -----> 0.13moles of HCl
cross multiply
24×0.13 =X×2
3.12. =2x
divide both sides by 2
x=3.12/2
x= 1.56g
which is still 1.5g as the option may have it

Anela

1 year ago

m= ?
mm=24
C= 0.5
V=260

m/mm = CV/1000

m/24 = 0.5×260/1000

= m/24 = 130/1000
CROSS MULTIPLY ✖️

1000m=3120
m =3.12

FROM THE EQUATION OF REACTION, 2 mols OF Hcl WAS INVOLVED.
Mg+2Hcl →MgCl2 + H2

SO; m= 3.12/2
=1.56

Macandcheese

1 year ago

no explanation