20g of NaHCO3 Treated with 22g of HCl Producing 12g of the residue. What is the amount of co2 Formed ?
NaHCO3 + HCl ==> NaCl + H2O + CO2
mols NaHCO3 = grams/molar mass = 20/84 = approx 0.24.
mols CO2 formed if NaHCO3 is the limiting reagent (LR)is
0.24 mols NaHCO3 x (1 mol CO2/1 mol NaHCO3) = 0.24 mols CO2.
mols HCl = 22/35.5 = approx 0.62.
mols CO2 formed if HCl is the LR is 0.6 x (1 mol CO2/1 mol HCl) = 0.6 mols CO2.
In LR problems, the SMALLER amount is ALWAYS the correct answer; therefore, 0.24 mols CO2 will be formed.
grams CO2 = 0.24 mol CO2 x molar mass CO2 = ? g is you want grams and not mols CO2.
NOTE: I have estimated these numbers; you should redo all of the calculations to get better numbers.
I want it in grams
"I want" -- ??
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Note that I showed how to do it in mols AND in grams.
To determine the amount of CO2 formed in the reaction between NaHCO3 and HCl, we can use stoichiometry. Let's calculate step by step:
Step 1: Write the balanced equation for the reaction:
NaHCO3 + HCl -> NaCl + H2O + CO2
It shows that 1 mole of NaHCO3 reacts with 1 mole of HCl to produce 1 mole of CO2.
Step 2: Calculate the number of moles of NaHCO3 and HCl used:
Molar Mass of NaHCO3 = Atomic Mass of Na + Atomic Mass of H + Atomic Mass of C + 3 × Atomic Mass of O
= 23 + 1 + 12 + 3 × 16
= 84 g/mol
Number of moles of NaHCO3 = Mass of NaHCO3 / Molar Mass of NaHCO3
= 20 g / 84 g/mol
≈ 0.2381 mol
Molar Mass of HCl = Atomic Mass of H + Atomic Mass of Cl
= 1 + 35.5
= 36.5 g/mol
Number of moles of HCl = Mass of HCl / Molar Mass of HCl
= 22 g / 36.5 g/mol
≈ 0.6027 mol
Step 3: Determine the limiting reagent:
To find the amount of CO2 formed, we need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and limits the amount of product formed.
From the balanced equation, we can see that 1 mole of NaHCO3 reacts with 1 mole of HCl to produce 1 mole of CO2. Therefore, the limiting reagent will be the one with the lesser number of moles, which is NaHCO3 in this case.
Step 4: Calculate the moles of CO2 formed:
Since NaHCO3 is the limiting reagent, the number of moles of CO2 formed will be the same as the number of moles of NaHCO3 used.
Number of moles of CO2 = Number of moles of NaHCO3
= 0.2381 mol
Step 5: Convert moles of CO2 to grams:
Molar Mass of CO2 = Atomic Mass of C + 2 × Atomic Mass of O
= 12 + 2 × 16
= 44 g/mol
Mass of CO2 = Number of moles of CO2 × Molar Mass of CO2
= 0.2381 mol × 44 g/mol
≈ 10.481 g
Therefore, the amount of CO2 formed in the reaction is approximately 10.481 grams.