Answer:
Mass of hydrogen produced = 2.1 g
Mass of excess reactant left = 25.2 g
Explanation:
Given data:
Mass of Mg = 50.0 g
Mass of HCl = 75.0 g
Mass of hydrogen produced = ?
Mass of excess reactant left = ?
Solution:
Chemical equation:
Mg + 2HCl → MgCl₂ + H₂
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 50 g/ 24 g/mol
Number of moles = 2.1 mol
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 75 g/ 36.5 g/mol
Number of moles = 2.1 mol
now we will compare the moles of hydrogen gas with both reactant.
Mg : H₂
1 : 1
2.1 : 2.1
HCl : H₂
2 : 1
2.1 : 1/2×2.1 = 1.05 mol
HCl is limiting reactant and will limit the yield of hydrogen gas.
Mass of hydrogen:
Mass = number of moles × molar mass
Mass= 1.05 mol ×2 g/mol
Mass = 2.1 g
Mg is present in excess.
Mass of Mg left:
HCl : Mg
2 : 1
2.1 : 1/2×2.1 = 1.05
Out of 2.1 moles of Mg 1.05 react with HCl.
Moles of Mg left = 2.1 mol - 1.05 mol = 1.05 mol
Mass of Mg left:
Mass = number of moles × molar mass
Mass = 1.05 mol × 24 g/mol
Mass = 25.2 g
B. 6.45 mol F2
The mole ratio here is 3 mol F2 for every 1 mol S. Multiply 2.15 mol S by 3 to get the mol of F2 needed.
