Question

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Answer 1

Answer:

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O. What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? And 25 ml of solutions contains Hcl = 27.375/1000 * 25 = 0.6844 g. 2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g).

Answer 2

Answer:

Moles of 100 ml of 0.5 M HCl is (0.5×100/1000) = 0.05 of HCl

∴ 0.05 of HCl are given to us.

∴ 1 mole of CaCO₃ is required for 2 mole of HCl reaction

∴ x mole of CaCO₃ is required for 0.05 mole of HCl reaction

∴ Given moles of  CaCO₃ is (0.05 ÷ 2) = 0.025

∴ Mass of CaCO₃ required for reaction is 100 × 0.025 = 2.5 gm

$Mass$ $of$ $CaCO_3$ $required$ $is$ $2.5 grams.$

∴ 1 mole of CaCO₃ give 1 mole of CO₂

∴ 0.025 moles of CaCO₃ give 0.025 mole of CO₂

∴ Mass of CO₂ produced is (0.025 × 44) = 1.1 gm

$Mass$ $of$ $CO_2$ $produced$ $is$ $1.1 grams.$