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
The empirical formula is Fe₃O₄.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of Fe to O.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio</u>¹ <u>×3</u>² <u>Integers</u>³
Fe 0.77 1 3 3
O 1.0 1.3 3.9 4
¹ To get the molar ratio, you divide each number of moles by the smallest number (0.77).
² If the ratio is not close to an integer, multiply by a number (in this case, 3) to get numbers that are close to integers.
³ Round off these numbers to integers (3 and 4).
The empirical formula is Fe₃O₄.
Answer:
The setup above is used to collect hydrogen gas
In the area of positive charge, neutral body gets negative charge for making a dipole whereas when negative charge bought near neutral one, it gets positive charge for the same reason.........making polarization
