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 )

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 =

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 =

volume =

volume = 0.0514 l

volume = 51.4 ml

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.