2.23 moles of propane react when 294 g of CO₂ is formed .
<h3>What is moles ?</h3>
Moles is a unit which is equal to the molar mass of an element.
A reaction is given
C₃H₈ +50₂ → 3CO₂ + 4H₂O
Grams of CO₂ formed = 294 gm
In moles = 294 /44 = 6.68 moles.
Let x be the moles of C₃H₈ is x
Mole ratio of CO₂ to C₃H₈ = 3 : 1
so
6.68 /x = 3/1
x = 6.68 /3 = 2.23 moles
Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .
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Answer:
from the 1st equation:
4NH3 4NO
4 *(68) 4*30
1216 X mass of NO = 536.5 g
from the 2nd Equation
2NO 2NO2
2*30 2* 46
536.5 x mass of NO2 = 822.6 grams
from the 3rd Equation
3NO2 2HNO3
3*(46) 2* (63)
822.6 X mass of nitric acid = 751.06 gram
b) % yields = ( 96.2%* 91.3% *91.4%)= 80.3%
Answer:
Mass of water = 73.08 g
Explanation:
Given data:
Mass of hydrogen = 35 g
Mass of oxygen = 65 g
Mass of water = ?
Solution:
First of all we will write the balanced chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass/ molar mass
Number of moles of hydrogen = 35 g/ 2 g/mol
Number of moles of hydrogen = 17.5 mol
Number of moles of oxygen = 65 g / 32 g/mol
Number of moles of oxygen = 2.03 moles
Now we compare the moles of water with moles hydrogen and oxygen.
H₂ : H₂O
2 : 2
17.5 : 17.5
O₂ : H₂O
1 : 2
2.03 : 2× 2.03 =4.06 mol
Number of moles of water produced by oxygen are less so oxygen is limitting reactant.
Mass of water:
Mass of water = number of moles × molar mass
Mass of water = 4.06 mol × 18 g/mol
Mass of water = 73.08 g
Answer is: [COCl₂] > [CO][Cl₂]
Chemical reaction: COCl₂(g) ⇄ CO(g) + Cl₂(g); Keq = 8.1 x 10⁻⁴.
Keq = [CO] · [Cl₂] / [COCl₂]; equilibrrium constant of chemical reaction.
[CO] · [Cl₂] / [COCl₂] = 0,00081.
Equilibrium product concentration is much more less than equilibrium concentration of reactant.
