The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.
Given chemical reaction is:
2B₂O₃ + 7C → B₄C + 6CO
From the stoichiometry of the reaction, it is clear that:
2 moles of B₂O₃ = produces 1 mole of B₄C
Now mass of B₄C will be calculated by using the below equation:
W = (n)(M), where
- n = moles = 1 mole
- M = molar mass = 55.25 g/mole
W = (1)(55.25) = 55.25 g
Hence required mass of B₄C is 55.25 grams.
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<span>i get 3.19x10^20 atoms
</span>
Explanation:
<em><u>2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2</u></em>
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The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
