The 6.
1. 0,005
5x10 3.
7.
2. 5,050
=
88104
8.
3. 0.0008 =
1 X 103
9.
4. 1,000 =
10
10
5. 1,000,000 =

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage =  77.77%

Explanation:

Step 1: 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

Step 2: 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

Step 3: 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 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: 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

Step 6: 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%