Mole ratios
Reactants and products
Type of reaction eg equilibrium
Enthalpy
Charges of ions
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer:
0.0308 mol
Explanation:
In order to convert from grams of any given substance to moles, we need to use its molar mass:
- Molar mass of KAI(SO₂)₂ = MM of K + MM of Al + (MM of S + 2*MM of O)*2
- Molar mass of KAI(SO₂)₂ = 194 g/mol
Now we <u>calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g</u>:
- 5.98 g ÷ 194 g/mol = 0.0308 mol
The density is 4 g/cm³ or 4000 kg/m³.
Density = mass/volume = 12 g/3 cm³ = 4 g/cm³
The measurement of 4 g/cm³ is already in <em>SI units</em>.
In SI <em>bas</em>e units,
Density = (4 g/1 cm³) × (1 kg/1000 g) × (100 cm/1 m)³ = 4000 kg/m³
<h3>
Answer:</h3>
Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.
<h3>
Explanation:</h3>
From the questions we are given;
4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(s)
- Moles of FeS₂ are 26.62 moles
- Moles of Oxygen, O₂ are 59.44 moles
We are supposed to determine the limiting and excess reactants;
- From the equation of the reaction given; 4 moles of FeS₂ required 11 moles of Oxygen gas.
Working with the amount of reactants given;
- 26.62 moles of FeS₂ will require 73.205 moles of O₂ and only 59.44 moles of O₂ are available.
- On the other hand 59.44 moles of O₂ requires 21.615 moles of FeS₂, and we are given 26.62 moles of FeS₂ which means FeS₂ is in excess.
Conclusion;
We can conclude that Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.
