In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. The distinction between physical and chemical change is not clear cut.
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g
C H O
mass 63.13 g 8.830 g 28.03 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.13/12 8.830/1 28.03/16
5.26 8.830 1.75
divide by the smallest number of moles
5.26/1.75 8.830/1.75 1.75/1.75
= 3.01 = 5.04 =1
rounded off to the nearest whole numbers
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O
Q2)
we have to next determine the molecular formula of the compound
molecular formula gives the actual composition of elements in the compound.
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
= 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
= 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄
Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = ![\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
