The mass of CO₂ gas = 4.312 g
<h3>Further explanation</h3>
Reaction
NaHCO₃ (s) + HCl (aq) → NaCl (aq) + H₂O (aq) + CO₂ (g)
Find limiting reactant
mol NaHCO₃

mol HCl

Limiting reactant : NaHCO₃ (smaller)
mol CO₂ = mol NaHCO₃ = 0.098
mass CO₂ :

Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:

Now, find moles of
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
Is it the dry lab/wet lab week 1 or ?
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