1.2 g of is treated with 100 mL of 1 M . Assuming complete reaction, the molar concentration of the — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Reaction: $Mg + H_2SO_4 \rightarrow MgSO_4 + H_2$. Initial moles of $Mg = 1.2 / 24 = 0.05$ mol. Initial moles of $H_2SO_4 = 100 \times 10^{-3} \times 1 = 0.1$ mol. Since they react 1:1, 0.05 moles of $H_2SO_4$ are consumed. Remaining moles of $H_2SO_4 = 0.1 - 0.05 = 0.05$ mol. Volume is 100 mL (0.1 L). Molarity = $0.05 / 0.1 = 0.5 M$.