Question

Baking powder is a mixture of sodium bicarbonate and an acid, often cream of tartar (made of tartaric acid, H2C4H5O6). These two components react when they are dissolved in water. A partial balanced equation for this reaction is shown below. Fill in the blank spaces with the appropriate compounds and/or coefficients: __NaHCO3(aq)+__H2C4H5O6(aq)-->______+_______+Na2C4H5O6(aq). A particular formulation of baking powder contains 2.13 g sodium bicarbonate and 4.67 g tartaric acid. How many millileters of gas will be released at STP?(R constant is 0.082 057 366 L atm/mol K)(If possible please try and help me with the net ionic equation too)

Answer 1

Balanced equation : 2 NaHCO₃(aq) + H₂C₄H₅O₆(aq) 2 CO₂(g) + 2 H₂O(l) + Na₂C₄H₅O₆(aq)

Case 1 : sodium bicarbonate NaHCO₃ is the limiting reactant

Given : mass NaHCO₃ reacted = 2.13 g

moles NaHCO₃ reacted = (mass NaHCO₃ reacted) / (molar mass NaHCO₃)

moles NaHCO₃ reacted = (2.13 g) / (84.0 g/mol)

moles NaHCO₃ reacted = 0.025355 mol

moles CO₂ formed = moles NaHCO₃ reacted

moles CO₂ formed = 0.025355 mol

Case 2 : tartaric acid H₂C₄H₅O₆ is the limiting reactant.

Given : mass H₂C₄H₅O₆ reacted = 4.67 g

moles H₂C₄H₅O₆ reacted = (mass H₂C₄H₅O₆ reacted) / (molar mass H₂C₄H₅O₆)

moles H₂C₄H₅O₆ reacted = (4.67 g) / (151.1 g/mol)

moles H₂C₄H₅O₆ reacted = 0.0309 mol

moles CO₂ formed = (moles H₂C₄H₅O₆ reacted) * (2 moles CO₂ / 1 mole H₂C₄H₅O₆)

moles CO₂ formed = (0.0309 mol) * (2 / 1)

moles CO₂ formed = 0.618 mol

Since less moles of CO₂ are formed in Case 1, therefore, sodium bicarbonate is the limiting reactant.

moles of CO₂ formed = 0.025355 mol

volume of CO₂ formed = (moles of CO₂ formed) * (molar volume at STP)

volume of CO₂ formed = (0.025355 mol) * (22.4 L/mol)

volume of CO₂ formed = 0.568 L

volume of CO₂ formed = 568 mL

Net ionic equation : HCO₃^- (aq) + H⁺ (aq) CO₂ (g) + H₂O (l)