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How many grams of sodium carbonate will react with 50 ml of 0.2 n HCl
Accepted Answer
To determine the mass of sodium carbonate required, we need to consider the balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl. First, we need to convert the volume and concentration of HCl to moles:
Volume of HCl = 50 mL = 0.05 L
Concentration of HCl = 0.2 N
Moles of HCl = Volume × Concentration = 0.05 L × 0.2 N = 0.01 moles
Since 2 moles of HCl react with 1 mole of Na2CO3, we can determine the moles of Na2CO3 required:
Moles of Na2CO3 = Moles of HCl / 2 = 0.01 moles / 2 = 0.005 moles
Finally, we can convert the moles of Na2CO3 to grams using its molar mass (105.99 g/mol):
Mass of Na2CO3 = Moles of Na2CO3 × Molar mass = 0.005 moles × 105.99 g/mol = 0.53 grams
Therefore, 0.53 grams of sodium carbonate will react with 50 ml of 0.2 N HCl.
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