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40.0g of K reacts with 40.0g of Cl_(2) to form KCl. Which is the limiting reagent? 2K(s)+Cl_(2)(g)arrow 2KCl(s) Cl_(2) KCI K 13 points

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40.0g of K reacts with 40.0g of Cl_(2) to form KCl. Which is the limiting reagent? 2K(s)+Cl_(2)(g)arrow 2KCl(s) Cl_(2) KCI K 13 points

40.0g of K reacts with 40.0g of Cl_(2) to form KCl. Which is the limiting reagent? 2K(s)+Cl_(2)(g)arrow 2KCl(s) Cl_(2) KCI K 13 points

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YasminProfessional · Tutor for 6 years

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### K

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## Step 1: Determine Molar Mass<br />Find the molar masses of each reactant:<br />- \(K\) (Potassium) has a molar mass of \(39.1 \, \text{g/mol}\)<br />- \(\mathrm{Cl}_{2}\) (Chlorine gas) has a molar mass of \(70.9 \, \text{g/mol}\)<br /><br />## Step 2: Calculate Moles of Each Reactant<br />Convert the given masses to moles:<br />\[ n_{\text{K}} = \frac{40.0 \, \text{g} \, \mathrm{K}}{39.1 \, \text{g/mol}} = 1.02 \, \text{mol} \]<br />\[ n_{\mathrm{Cl}_{2}} = \frac{40.0 \, \text{g} \, \mathrm{Cl}_{2}}{70.9 \, \text{g/mol}} = 0.564 \, \text{mol} \]<br /><br />## Step 3: Determine Stoichiometric Ratios<br />From the balanced equation:<br />\(2 \, \text{K} + 1 \, \mathrm{Cl}_{2} \rightarrow 2 \, \mathrm{KCl}\)<br /><br />The ratio required by the reaction is:<br />\[ \frac{2 \, \text{mol} \, \text{K}}{1 \, \text{mol} \, \mathrm{Cl}_{2}} \]<br /><br />## Step 4: Compare Available Ratios<br />Compare the mole ratio of the available reactants to the stoichiometric ratio:<br />\[ 1 \, \text{mol} \, \mathrm{Cl}_{2} \text{ requires } 2 \, \text{mol} \, \text{K} \]<br /><br />Calculate the limiting reagent:<br />\[ \text{Required moles of } \text{K for available } \mathrm{Cl}_{2} = 0.564 \, \text{mol} \times 2 = 1.128 \, \text{mol} \]<br />Since we only have \(1.02 \, \text{mol} \, \text{K}\) but need \(1.128 \, \text{mol}\), \(\text{K}\) is less than required, making \(\text{K}\) the limiting reagent.
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