Calculate the weight (in mg) of borax Na2B4O7.10H20 required to react with 25.0 mL if 0.100 M HCI

Question

No. moles HCl consumed = __________ moles
No. moles borax required = __________ moles
Molar mass of borax = __________ g mol-1
(Na2B4O7.10H2O)
Weight borax required = __________ g
= __________ mg

My calculations:
2 HCl + Na2B4O7.10 H2O → H2B4O7 + 2 NaCl + 10 H2O
Number of moles = Volume in liters * Molarity = 0.025 * 0.1 = 0.0025
The ratio of HCl to Na2B4O7.10 H2O is 2 to 1
No. of moles borax required = ½ * 0.0025 = 0.00125

Molar mass of borax = 2 * 23 + 4 * 10.8 + 7 * 16 + 10 *18 = 381.2 grams
Mass of borax = 0.00125 * 381.2
Weight borax required = 0.00125 * 381.2 * 9.8

Answer 1

All of these are ok except for the last one. See below
My calculations:
2 HCl + Na2B4O7.10 H2O → H2B4O7 + 2 NaCl + 10 H2O
Number of moles = Volume in liters * Molarity = 0.025 * 0.1 = 0.0025
The ratio of HCl to Na2B4O7.10 H2O is 2 to 1
No. of moles borax required = ½ * 0.0025 = 0.00125

Molar mass of borax = 2 * 23 + 4 * 10.8 + 7 * 16 + 10 *18 = 381.2 grams
Mass of borax = 0.00125 * 381.2
Weight borax required = 0.00125 * 381.2 * 9.8
g borax needed = mols x molar mass and that is 0.00125 x 381.2 = ?
Then convert to kg and that x 9.8 = ?