Answer:
126.0g of water were initially present
Explanation:
The electrolysis of water occurs as follows:
2H₂O(l) ⇄ 2H₂(g) + O₂(g)
<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>
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To find the mass of water we need to determine moles of oxygen and hydrogen, thus:
<em>Moles Hydrogen:</em>
14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂
<em>Moles Oxygen:</em>
112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂
Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:
7 moles H₂O * (18g / mol) =
<h3>126.0g of water were initially present</h3>
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced. - online resource
Explanation: if not im sorry
B. Theory
The hypothesis that stand the test of time (often tested and never rejected) is called theory. A theory is supported by a great dealcof evidence.
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