Question

2HCl + Mg mc031-1.jpg MgCl2 + H2

If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?

Answer 1

Percentage Yield = mass of Actual Yield/mass of Theoretical Yield*100%

Balanced equation is: Mg+2HCl=MgCl2 + H2Showing that if Mg is not a limiting factor then 2 moles of HCl on complete reaction liberate 1 mole of Hydrogen
1 mole of HCl=35.5g, 40g=xX= 40/35.5= 1.127mole2 moles of HCl = 1 moles of Hydrogen1.127=XX=1.127/2 = 0.56
Hope this helped :)

Answer 2

1.10 g H₂

Step-by-step explanation:

The reaction is

2HCl + Mg → MgCl₂ + H₂

An excess of magnesium metal means that the limiting reactant is HCl, so we use the mass of HCl given by the problem to calculate how many grams of H₂ are produced.

In order to make this calculation we need to keep in mind the molar mass of HCl, of H₂, and the reaction stoichiometry:

40.0 g HCl *
`(1molHCl)/(36.5g) *(1molH_(2))/(2molHCl)*(2g)/(1molH_(2))=` 1.10 g H₂