Respuesta :
The molarity of HCl is calculated as below
find the moles of mg used
moles = mass /molar mass
= 2.00g/24 g/mol=0.083 moles
write the reacting equation
Mg(s) +2HCl (aq)= MgCl2(aq) + H2(g)
by use of mole ratio between Mg to HCl which is 1 :2 the moles of HCl is therefore
= 0.083 x2 =0.166 moles
molarity of HCl = moles of HCl/ volume in dm^3
that is 0.166/50 x1000= 3.3 M of HCl
find the moles of mg used
moles = mass /molar mass
= 2.00g/24 g/mol=0.083 moles
write the reacting equation
Mg(s) +2HCl (aq)= MgCl2(aq) + H2(g)
by use of mole ratio between Mg to HCl which is 1 :2 the moles of HCl is therefore
= 0.083 x2 =0.166 moles
molarity of HCl = moles of HCl/ volume in dm^3
that is 0.166/50 x1000= 3.3 M of HCl
the balanced equation for the reaction between Mg and HCl is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of Mg to HCl is 1:2
number of moles of Mg reacted - 2.00 x 10⁻³ g / 24 g/mol = 0.0833 x 10⁻³ mol
according to molar ratio
number of HCl moles reacted - 0.0833 x 10⁻³ x 2 mol = 0.167 x 10⁻³ mol
number of HCl moles in 50.0 mL - 0.167 x 10⁻³ mol
therefore number of HCl moles in 1000 mL - 0.167 x 10⁻³ mol / 0.0500 L
molarity of HCl - 0.00334 M
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of Mg to HCl is 1:2
number of moles of Mg reacted - 2.00 x 10⁻³ g / 24 g/mol = 0.0833 x 10⁻³ mol
according to molar ratio
number of HCl moles reacted - 0.0833 x 10⁻³ x 2 mol = 0.167 x 10⁻³ mol
number of HCl moles in 50.0 mL - 0.167 x 10⁻³ mol
therefore number of HCl moles in 1000 mL - 0.167 x 10⁻³ mol / 0.0500 L
molarity of HCl - 0.00334 M