Answer:
Explanation:
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In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
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Explanation:
Below is an attachment containing the solution.
Mass of Li₂O= 719.809 g
<h3>Further explanation</h3>
Given
1,060 g of CO₂
Required
Mass of Li₂O
Solution
Reaction
<em>Possible compound : Li₂O </em>
Li₂O(aq) + CO₂(g) → Li₂CO₃(aq)
mol CO₂(MW=44 g/mol) :
= 1,060 g : 44 g/mol
= 24.09
From the equation, mol ratio of LiO : CO₂ = 1 : 1, so mol Li₂O : 24.09
Mass Of Li₂O(MW=29.88 g/mol) :
= mol x MW
= 24.09 x 29.88
= 719.809 g
<span>A scientific Law is a principle that describes the behavior of a natural phenomenon.
so it is A.</span>