When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃
Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
Empirical formula is the simplest ratio of whole numbers of components in a compound
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements simplified ratio
C : H C : H
ethyne 2:2 1:1
ethene 2:4 1:2
ethane 2:6 1:3
methane 1:4 1:4
the only compound where the actual ratio is equal to the simplified ratio is methane
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄
The SI unit for measuring distance is the
meter.
I attached a table of SI measurements for you :)
Hope this helped!