M(H₂O) = 97,2 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 97,2 g ÷ 18 g/mol.
n(H₂O) = 5,4 mol.
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 5,4 mol · 6,023·10²³ 1/mol.
N(H₂O) = 3,25·10²⁴ molecules of water.
n - amount of substance.
Na - Avogadro number.
Answer:
Explanation:
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The question is incomplete. The complete question is :
C. Balance these fossil-fuel combustion reactions. (1 point)
C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat
CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
Solution :
C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat
When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.
CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat
When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.
C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat
When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.
C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat
When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.
The answer is A) Electrons are exchanged.
All reactions that involve molecular oxygen, such as combustion and corrosion, are electron transfer reactions. This includes the rusting of iron.
The product will not be affected by the addition of twice as much Na₂CO₃.
<h3>What is Limiting reagent in stoichiometry ?</h3>
- The maximum quantity of the end product determined by a balanced chemical equation is known as the Stoichiometry.
- The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced, and the one which remains unconsumed after the final reaction is in Excess.
- Calculate the moles of each reactant present and contrast it with the mole ratio of the reactants in the balanced equation to determine which reactant is the limiting one.
Here,taking the stoichiometry into consideration, we find that the reaction happens with 1:1 ratio; so, adding twice the amount of Na₂CO₃ will lead to its excess making the other the limiting reactant, hence, it would not affect the yield of the product.
To know more about the Limiting reactant, refer to:
brainly.com/question/14222359
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