Question

Nahco3 can be used to neutralize hydrogen ion, h+. if nahco3 can be purchased at a cost of $9.20/kg, how much will it cost to neutralize 1.0 mole of h+?

Answer 1

NaHCO₃ + H⁺ = Na⁺ + H₂O + CO₂

n(NaHCO₃)=n(H⁺)

m(NaHCO₃)=M(NaHCO₃)n(NaHCO₃)=M(NaHCO₃)n(H⁺)

p=$m(NaHCO₃)=$M(NaHCO₃)n(H⁺)

p=9.20$/kg×84.0kg/kmol×10⁻³kmol=0.7728$

p≈0.77$

Answer 2

`Cost=\$0.77`

Step-by-step explanation:

Hello,

At first, the ionic neutralization chemical reaction turns out into:

`NaHCO_3+H^+-->Na^++H_2O+CO_2`

In such a way, we develop the stoichiometry calculation to obtain the kilograms of sodium bicarbonate:

`m_(NaHCO_3)=1.0 molH^+*(1molNaHCO_3)/(1molH^+)*(84gNaHCO_3)/(1molNaHCO_3) *(1kgNaHCO_3)/(1000gNaHCO_3) \\m_(NaHCO_3)=0.084kgNaHCO_3`

Now, we compute the cost by considering the neutralized kilograms of sodium bicarbonate as shown below:

`Cost=\$9.20/kg*0.084kg\\Cost=\$0.77`

Best regards.