What volume (in mL) of a 0.150 M HNO 3 solution will completely react with 35.7 mL of a 0.108 M Na 2 CO 3 solution according to this balanced chemical equation?
Na 2 CO 3 ( a q ) + 2 HNO 3 ( a q ) → 2 NaNO 3 ( a q ) + CO 2 ( g ) + H 2 O ( l )

Question

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Respuesta :

joseaaronlara joseaaronlara · Answer 1 · 2019-12-11T19:50:53+01:00

Answer:

The answer to your question is 51.4 ml

Explanation:

Data

HNO₃ = 0.150 M; Volume = ?

Na₂CO₃ = 0.108 M; Volume = 35.7 ml

Process

1.- Calculate number of moles of Na₂CO₃

Molarity = [tex]\frac{moles}{volume}[/tex]

moles = Molarity x volume

moles = 0.108 x 0.0357

moles = 0.00386

2.- From the equation we know that 1 mol of Na₂CO₃ reacts with 2 moles of HNO₃.

1 mol of Na₂CO₃ ------------- 2 moles of HNO₃

0.00386 mol of Na₂CO₃ ------ x

x moles of HNO₃ = (0.00386 x 2) / 1

moles of HNO₃ = 0.00771

3.- Calculate the volume of HNO₃ used

volume = [tex]\frac{moles}{molarity}[/tex]

volume = [tex]\frac{0.00771}{0.150}[/tex]

volume = 0.0514 l

volume = 51.4 ml