Question

what volume of 0.250 M HNO3 (nitric acid) reacts with 44.8ml of 0.150 M na2co3 ( sodium carbonate) in the following equation?

2HNO3(aq)+Na2co3(aq)-------2nano3(aq)+h20+co2

Answer 1

Answer – Given, [HNO₃] = 0.250 M , [Na₂CO₃] = 0.150 M , volume = 44.8 mL

Reaction –

Na₂CO₃ + 2HNO₃ -----> 2 NaNO₃ + CO₂ + H₂O

We need to calculate the moles of Na₂CO₃

So, moles of Na₂CO₃ = 0.150 M * 0.0448 L

                                   = 0.00672 moles

From the balanced reaction –

1 moles of Na₂CO₃ = 2 moles of HNO₃

So, 0.00672 moles Na₂CO₃ = ?

= 0.0134 moles of HNO₃

Now we know,

Molarity = moles / L

So, volume (L) = moles / molarity

                        = 0.0134 moles / 0.250 M

                   = 0.0538 L

We know,

1 L = 1000 mL

So, 0.0538 L = ?

= 53.8 mL

53.8 mL of 0.250 M HNO₃ (nitric acid) reacts with 44.8ml of 0.150 M Na₂CO₃.