Question

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 )

Answer 1

The answer to your question is 51.4 ml

Step-by-step explanation:

Data

HNO₃ = 0.150 M; Volume = ?

Na₂CO₃ = 0.108 M; Volume = 35.7 ml

Process

1.- Calculate number of moles of Na₂CO₃

Molarity =
`(moles)/(volume)`

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 =
`(moles)/(molarity)`

volume =
`(0.00771)/(0.150)`

volume = 0.0514 l

volume = 51.4 ml