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Consider the reaction: MnO₂ + 4 HCl → MnCl₂ + Cl₂ + 2H₂O If 0.86 mole of MnO₂ and 48.2 g of HCl react, which reagent will be used up first? How many grams of Cl₂ will be produced?

User Jack Dre
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Final answer:

The limiting reagent in the chemical reaction between MnO₂ and HCl is HCl. Based on stoichiometry, 1.322 moles of HCl will produce 0.3305 moles (23.44 grams) of chlorine gas (Cl₂).

Step-by-step explanation:

The student's question concerns a chemical reaction where manganese dioxide (MnO₂) reacts with hydrochloric acid (HCl) to produce manganese(II) chloride (MnCl₂), chlorine gas (Cl₂), and water (H₂O).

To determine the limiting reagent and the amount of Cl₂ produced, we need to compare mole ratios using the balanced chemical equation. Using molar masses, we calculate the number of moles of HCl to be 48.2 g / 36.46 g/mol = 1.322 moles.

The reaction requires a 4:1 mole ratio of HCl to MnO₂, thus for 0.86 moles of MnO₂ we need 0.86 x 4 = 3.44 moles of HCl. Since we only have 1.322 moles of HCl, it will be the limiting reagent.

The mole ratio of HCl to Cl₂ in the balanced equation is 4:1, so the amount of Cl₂ produced from 1.322 moles of HCl will be 1.322 moles / 4 = 0.3305 moles. To find the mass of Cl₂ produced, we multiply the moles by the molar mass of Cl₂ (0.3305 moles x 70.90 g/mol = 23.44 grams).

User Rodrigo Ehlers
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