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WAEC SSCE - Chemistry - 2014 (Objective)

Pregunta 1 Informe

Consider the reaction represented by the following equation: Na2CO3(aq) + 2HCl(aq) \(\longrightarrow\) 2NaCl(aq) + H2O(l) + CO2(g) . What volume of 0.02 mol dm-3 Na2CO3(aq) would be required to completely neutralize 40 cm3 of 0.10 mol dm-3 HCl(aq)?
Detalles de la respuesta
The balanced chemical equation for the reaction between Na2CO3(aq) and HCl(aq) is: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) From the equation, we can see that 1 mole of Na2CO3(aq) reacts with 2 moles of HCl(aq) to produce 2 moles of NaCl(aq), 1 mole of H2O(l), and 1 mole of CO2(g). Therefore, the number of moles of HCl(aq) that reacted can be calculated as: number of moles = concentration x volume number of moles = 0.10 mol dm-3 x 0.040 dm3 number of moles = 0.004 mol From the balanced equation, we know that 1 mole of Na2CO3(aq) reacts with 2 moles of HCl(aq). Therefore, the number of moles of Na2CO3(aq) required to completely neutralize the HCl(aq) can be calculated as: number of moles = 0.5 x number of moles of HCl(aq) number of moles = 0.5 x 0.004 number of moles = 0.002 Finally, we can use the concentration and number of moles to calculate the required volume of Na2CO3(aq) as: volume = number of moles / concentration volume = 0.002 mol / 0.02 mol dm-3 volume = 0.1 dm3 = 100 cm3 Therefore, the correct answer is 100 cm3.
Lee la nota de la lección en Stoichiometry And Chemical Reactions (WAEC)
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