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

Answer:

$Cost=\ 0.77$

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:

$m_{NaHCO_3}=1.0 molH^+*\frac{1molNaHCO_3}{1molH^+}*\frac{84gNaHCO_3}{1molNaHCO_3} *\frac{1kgNaHCO_3}{1000gNaHCO_3} \\m_{NaHCO_3}=0.084kgNaHCO_3$

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.