Help with chemistry homework (stoichiometry)!

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I need help with number 8 (you can see it on the picture above). I have the answers written in from the teacher, however, I still can't figure out the process involved to get the answers. Please explain at least 8a step by step, and hopefully I can do the rest myself!

8a.
Calculate the molar mass of K2O and the molar mass of BaBr2 from the periodic table.
molar mass of K2O = 2(39.098) + 15.9994 = 94.1954 g K2O
molar mass of BaBr2 = 137.33 + 2(79.904) = 297.138 g BaBr2
Then use stoichiometry starting with your given:
35.0 g K2O * (1 mol K2O / 94.1954 g K2O) * (1 mol BaBr2 / 1 mol K2O) * (297.138 g BaBr2 / 1 mol BaBr2)
= 110.4069838 g BaBr2
With significant figures, that's 1.10 x 10^2 g BaBr2

Here's what's going on when you do the calculations
1. start with given (35.0 g K2O)
2. convert given mass to moles by using molar mass (multiply by 1 mol K2O / 94.1954 g K2O)
3. use the mole-to-mole ratio given by the balanced equation to get from K2O to BaBr2 (multiply by 1 mol BaBr2 / 1 mol K2O because in the reaction, you need one mole of BaBr2 for every one mole of K2O)
4. convert the moles of BaBr2 by multiplying it to its molar mass (multiply 297.138 g BaBr2 / 1 mol BaBr2)

BaBr2 + K2O >>>>BaO + 2KBr

8a may be wrong

you want grams of BaBr2 to react with 35.0 grams K2O

grams K2O x 1 x 1 moles A x mass weight of BaBr2 = grams BaBr2
1 mass weight of A 1 moles B 1

35 grams x 1 x 1mole BaBr2 x 297.1**
1 183.4* 1mole K2O 1

*mass weight of 2 Kr atoms and 1 O atoms
**mass weight of 1 Ba atom and 2 Br atoms
= 56.7 g BaBr₂ not 110g
but...........
grams BaBr2 x 1 x 1 moles K2O x MW K2O = grams K2O
1 MW BaBr2 1 moles BaBr2 1

110 grams x 1 x 1mole BaBr2 x 183.4** = 67.9
1 297.1* 1mole K2O 1

110 grams of BaBr2 yields 67.9 g K2O not 56.7, so working backwards did not confirm
the answer

if you plug in the answer your teacher gave
if you do the formula using the 67.9 K2O and not 56.7 you get 110g BaBr2