What volume of 3.00 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M Na2CO3?

0.0500L of 3.00M HCl are needed to react completely

Explanation:

Hydrochloric acid HCl reacts with sodium carbonate, Na2CO3, as follows:

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂

Where 2 moles of HCl reacts per mole of Na₂CO₃

To find the volume of 3.00M HCl needed to react completely we need first to calculate the moles of Na₂CO₃ added and, with this chemical equation we can solve for moles of HCl as follows:

Moles Na₂CO₃:

0.750L * (0.100moles Na₂CO₃ / L) = 0.0750moles

Moles HCl:

0.0750moles Na₂CO₃ * (2 moles HCl / 1 mole of Na₂CO₃) = 0.150 moles of HCl

Volume of 3.00M HCl:

0.150 moles of HCl * (1L / 3.00moles) =

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