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

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

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 = ?

No. moles HCl consumed = 0.0025 moles

No. moles borax required = 0.00125 moles
Molar mass of borax (Na2B4O7.10H2O) = 381.2 g/mol
Weight borax required = No. of moles * Molar mass of borax = 0.00125 * 381.2 = 476.5 grams
Weight borax required = 476.5 grams = 476500 mg

To calculate the weight of borax required to react with 25.0 mL of 0.100 M HCl, you need to follow these steps:

1. Calculate the number of moles of HCl consumed:
Number of moles = Volume in liters * Molarity
Given that the volume is 25.0 mL (which is equivalent to 0.025 L) and the molarity is 0.100 M, we can calculate:
Number of moles HCl consumed = 0.025 L * 0.100 mol/L = 0.0025 moles

2. Determine the mole ratio between HCl and borax (Na2B4O7.10H2O). The balanced equation shows that 2 moles of HCl react with 1 mole of borax.

3. Calculate the number of moles of borax required:
The ratio of HCl to borax is 2:1. Therefore, the number of moles of borax required will be half of the number of moles of HCl consumed:
Number of moles borax required = 0.0025 moles * 1/2 = 0.00125 moles

4. Find the molar mass of borax (Na2B4O7.10H2O) using the atomic masses of the elements:
Na: 2(23) = 46 g/mol
B: 4(10.8) = 43.2 g/mol
O: 7(16) = 112 g/mol
H: 10(1) = 10 g/mol
Total molar mass = 46 + 43.2 + 112 + 10 = 211.2 g/mol

5. Calculate the weight of borax required in grams:
Weight borax required = Number of moles borax required * Molar mass of borax
Weight borax required = 0.00125 moles * 211.2 g/mol = 0.264 g

6. Convert the weight to milligrams:
Weight borax required = 0.264 g * 1000 mg/g = 264 mg

Therefore, the weight of borax required to react with 25.0 mL of 0.100 M HCl is 264 mg.