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How many milliliters of 0.200 m HCl would be needed to react exactly with 12.0 g of solid NaHCO3? NaHCO3 (s) + HCl(aq) = NaCl (aq) + CO2(g) + H2O(l)
Accepted Answer
Here's how to calculate the volume of HCl needed:
1. Balance the equation: The given equation is already balanced.
2. Convert grams of NaHCO3 to moles:
Molar mass of NaHCO3 = 84.01 g/mol
Moles of NaHCO3 = 12.0 g / 84.01 g/mol = 0.143 mol
3. Use stoichiometry to find moles of HCl:
From the balanced equation, 1 mole of NaHCO3 reacts with 1 mole of HCl.
Moles of HCl = 0.143 mol NaHCO3
(1 mol HCl / 1 mol NaHCO3) = 0.143 mol HCl
4. Calculate volume of HCl:
Molarity = moles / volume
Volume = moles / molarity
Volume = 0.143 mol / 0.200 M = 0.715 L
5. Convert liters to milliliters:
Volume = 0.715 L
1000 mL/L = 715 mL
Therefore, you would need 715 mL of 0.200 M HCl to react exactly with 12.0 g of NaHCO3.
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