If a reaction mixture contains 24 g of Mg and 32 g of O2 , what is the limiting reactant?
1 answer:
This problem is providing the mass of both magnesium metal and oxygen gas and involved in a chemical reaction and asks for the limiting reactant. At the end, it turns out to be identified as magnesium.
<h3>Stoichiometry</h3>In chemistry, stoichiometry is a widely-used tool we use in order to relate the mass and moles of different chemical substances involved in a chemical reaction. Thus, we consider the following chemical equation between magnesium and oxygen to produce magnesium oxide.
However, when the mass of the both of the reactants is given, one must identify the limiting reactant as the one producing the least of the moles of the product, which means we can use the given grams of the both of the reactants, their molar masses and mole ratios with the product to obtain the aforementioned:
Thus, we can evidence how 24 g of magnesium produce the least of the moles of magnesium oxide, fact validating the magnesium as the limiting reactant and the oxygen as the excess one.
Learn more about stoichiometry: brainly.com/question/9743981
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Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:
Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M