Answer:
0.009725 moles of H2C2O4
0.009725 moles CaCO3
Mass percentage = 77.77%
Explanation:
<u>Step 1</u>: The balanced equation
2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O
We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O
<u>Step 2</u>: Calculate moles of MnO4-
Molarity = Moles/volume
Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-
Moles of Mno4- = 0.1092M * 35.62 *10^-3 L
Moles of MnO4- = 0.00389 moles
<u>Step 3</u>: Calculate moles of H2C2O4
Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-
then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = <u>0.009725 moles of H2C2O4</u>
<u />
<u>Step 4:</u> Calculate moles of CaCO3
moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3
<u>Step 5</u>: Calculate mass of CaCO3
Molar mass of CaCO3 = 100.09 g/mole
Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3
Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g
<u>Step 6</u>: Calculate percentage by weight of CaCO3
Mass of CaCO3 = 0.9734g
Mass of original sample = 1.2516g
Mass percentage = 0.9734/1.2516 *100% = 77.77%
