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012 3 2 In the reaction, 575g of sodium was added to 7.0g of chlorine. Explain why chlorine is the limiting reactant. You must show your working. __

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012 3 2
In the reaction, 575g of sodium was added to 7.0g of chlorine.
Explain why chlorine is the limiting reactant.
You must show your working.
__

012 3 2 In the reaction, 575g of sodium was added to 7.0g of chlorine. Explain why chlorine is the limiting reactant. You must show your working. __

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UwennaExpert · Tutor for 3 years

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The limiting reactant in this reaction, therefore, is chlorine.

Explain

The limiting reactant is the reactant that completely gets consumed first and determines when the reaction will stop. In this case, we have been given amounts of sodium and chlorine. We will find the limiting reactant by comparing the ratio of the moles we have to the ratio required as per the reaction.<br /><br />1. First, convert the given masses of the reactants to moles. The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of chlorine (Cl2) is 70.90 g/mol.<br /><br /> So, the moles of sodium = 5.75 g / 22.99 g/mol ≈ 0.25 moles.<br /> <br /> And, for chlorine, moles of chlorine = 7.0 g / 70.90 g/mol ≈ 0.099 moles.<br /><br />2. For the reaction, the balanced chemical equation is: 2Na + Cl2 → 2NaCl, meaning that 2 moles of Na react with 1 mole of Cl2.<br /><br />3. Given the present amounts, for stoichiometry 2:1 required by the reaction for Na:Cl2, if we cross multiply our actual amounts with the stoichiometry, we can see we don't have enough Cl2 (chlorine) for all the Na (sodium): <br /> <br /> Na : Cl2 ≈ 0.25 actual mole ratio : 0.099 actual mole ratio ≠ 2:1 stoichiometric mole ratio,<br /> <br /> Here, chlorine will be deficient in comparison to sodium, given our stoichiometric ratio, and thereby will be used up before sodium.
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