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In the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of a 1.0 M HCl in an excess of Mg?

Respuesta :

Willchemistry
HCl is limiting reactant, Mg in excess therefore:

number of moles HCl:

75.0 mL in liters: 75.0 / 1000 => 0.075 L

Molarity HCl = 1.0 M

n = M x V

n = 1.0 x 0.075 

n = 0.075 moles of HCl

Mg(s) + 2 HCl (aq) = H2 (g) + MgCl2 (aq)

2 moles HCl  -------------------- 1 mole H2
0.075 moles of HCl ------------ ( moles H2)

moles H2 = 0.075 x 1 / 2

moles H2 = 0.075 / 2

= 0.0375 moles of H2

hope this helps!
Number of moles Hydrochloric Acid (HCL) = 0.0750*1.0 
                                                                    = 0.0750 
Hydrochloric Acid and Hydrogen have a ratio of 2 is to 1 

Number of moles Hydrogen = 0.0750/2
                                             = 0.0375

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