Answer:
<span>23.6
g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from
15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because
it gets used up and only makes 10.7 g of CO2. </span>
Explanation:
1) Balanced chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O
3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol
4) Convert the reactant masses to number of moles, using the formula
number of moles = mass in grams / molar mass
CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
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O₂: 15.6 g / 32.0 g/mol = 0.4875 moles
5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂
Which is what the first part of the answer says.
6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.
Which is what the second part of the answer says.
7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.
Which is what the chosen answer says.
8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.
As you go from left to right and down to up on the periodic table, atomic radius decreases. This would make helium the smallest and francium the largest.
Isn’t this app cool people are willing to help
Answer:
The total number of molecules remains the same
