Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
- Mg: 1 mole
- HCl: 2 moles
- MgCl₂: 1 mole
- H₂: 1 mole
Being:
- Mg: 24. 31 g/mole
- H: 1 g/mole
- Cl: 35.45 g/mole
the molar mass of the compounds participating in the reaction is:
- Mg: 24.31 g/mole
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
- H₂: 2*1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
- Mg: 1 mole* 24.31 g/mole= 24.31 g
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
- H₂: 1 mole* 2 g/mole= 2 g
Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]
moles of H₂= 0.134
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
