Answer:
Carbon monoxide is a very important industrial compound. In the form of producer gas or water gas.
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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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The number of mole will be 65.81 mole.
An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.
Number of ideal gas can be calculated by using the formula:
PV = nRT
where, p is pressure, n is number of mole, R is gas constant and T is temperature.
Given data:
V= 1750 
= 1750 L
P = 125,000 p = 1.2 atm
R = 0.082 L /mole kelvin
T = 273+127 = 400 K
Now, put the value of given data in above equation.
1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin x 400K
n = 65.81 mole.
Therefore, the number of mole will be 65.81 mole
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Answer:107.1 g, 124.1 g
Explanation:
The equation of the reaction is;
Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)
Hence;
For Al2S3
Number of moles= reacting mass/molar mass
Number of moles = 158g/150gmol-1 =1.05 moles
If 1 mole of Al2S3 yields 3 moles of H2S
1.05 moles of Al2S will yield
1.05 × 3/1 = 3.15 moles
Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g
For water
Number of moles of water = 131g/18gmol-1= 7.3 moles
6 moles of water yields 3 moles of H2S
7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S
3.65 moles × 34 gmol-1 =124.1 g
Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
