question archive Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4
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Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit.
Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit.
ANSWER
Moles of oxalic acid ued = mass used/molar mass
Molar mass of oxalic acid= 90.03 g/mol
Moles of oxalic acid ued = 0.20 g ÷ 90.03 g/mol
=0.0022moles
Equation for the reaction is;
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
The equation shows that the reacting mole ratio of H2C2O4 : NaOH is 1:2
The moles of the acid used is;
H2C2O4 : NaOH
1 : 2
0.0022 : X
X=(2×0.0022)÷1
= 0.0044 moles
Volume of NaOH used in the reaction is;
0.050 moles are present in 1000 mL
0.0044 moles are present in X mL
X =(0.0044 moles × 1000 mL)÷ 0.050 moles
= 88 mL
Step-by-step explanation
Oxalic acid is a solid organic acid that can donate two protons and has the molecular formula H2C2O4. How many milliliters of 0.050 M NaOH are needed to neutralize both acidic protons in 0.20 g of oxalic acid? You must show all your calculations to receive credit.
ANSWER
Moles of oxalic acid ued = mass used/molar mass
Molar mass of oxalic acid= 90.03 g/mol
Moles of oxalic acid ued = 0.20 g ÷ 90.03 g/mol
=0.0022moles
Equation for the reaction between oxalic acid and sodium hydroxide is used to calculate moles of acid used. The equation is;
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
The equation shows that the reacting mole ratio of H2C2O4 : NaOH is 1:2
The moles of the acid used is;
H2C2O4 : NaOH
1 : 2
0.0022 : X
X=(2×0.0022)÷1
= 0.0044 moles
Volume of NaOH used in the reaction is;
0.050 moles are present in 1000 mL
0.0044 moles are present in X mL
X =(0.0044 moles × 1000 mL)÷ 0.050 moles
= 88 mL