Balance Chemical Equation for combustion of Propane is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
According to equation,
1 mole of C₃H₈ on combustion gives = 4 moles of H₂O
So,
5 moles of C₃H₈ on combustion will give = X moles of H₂O
Solving for X,
X = (5 mol × 4 mol) ÷ 1 mole
X = 20 moles of H₂O
Calculating number of molecules for 20 moles of H₂O,
As,
1 mole of H₂O contains = 6.022 × 10²³ molecules
So,
20 moles of H₂O will contain = X molecules
Solving for X,
X = (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol
X = 1.20 ×10²⁵ Molecules of H₂O
A decomposition reaction occurs when one reactant breaks down into two or more products. It can be represented by the general equation:
AB → A + B
In this equation, AB represents the reactant that begins the reaction, and A and B represent the products of the reaction. The arrow shows the direction in which the reaction occurs.
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
</span>
Answer:
No more than 0.1 mL of hydrochloricton acid
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
