How many moles of carbon dioxide are produced when 12.1 g of HF are reacted with an excess amount of sodium carbonate? Na2CO3 (aq) + 2 HF (aq) → H2O (ℓ) + CO2 (g) + 2 NaF (aq)

0.3025 moles of CO₂  are produced

Explanation:

Reaction is: Na₂CO₃ (aq) + 2 HF (aq) → H₂O (l) + CO₂ (g) + 2 NaF (aq)

It is correctly ballanced.

If the given information stays that the excess amount is on the sodium carbonate, the HF is the limiting reagent.

We see that stoichiometry is 2:1. Moles of produced CO2, will be the half, than the original amount, we have.

Firste step: We convert the mass to moles

12.1 g . 1mol / 20g = 0.605 moles of HF

2 moles of HF can produce 1 mol of CO₂

0.605 moles of HF will produce (0.605 . 1) /2 = 0.3025 moles of CO₂

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg

the answer is 1 i believe