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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

i think it's letter a.