0.125 mol Ca
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Step 1: Define</u>
5.00 g Ca
<u>Step 2: Identify Conversions</u>
Molar mass of Ca - 40.08 g/mol
<u>Step 3: Convert</u>
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.12475 mol Ca ≈ 0.125 mol Ca
Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of
Thus 0.59 moles of give =
of
Mass of
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer:
Explanation:
Hello there!
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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