Question

When 35.47 g of sodium hydroxide react with boric acid (H3BO3), how many moles of sodium borate will be produced?

Answer 1

Answer:

5.83 g

Explanation:

First, you must start with a balanced equation so you can see the mole ratios.

NaOH + H₃BO₃ --> NaBO₂ + 2H₂O

You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio

Now you must calculate how many moles of NaOH 35.47 g equals.

Na = 22.99 amu

O = 15.99 amu

H = 1.008 amu

NaOH = 39.997 amu

35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH

Since it's a 1:1 ratio, the same number of moles of NaBO₂ is created. Now you must convert moles to grams.

Na = 22.9 amu

B = 10.81 amu

2 O = 31.998 amu

NaBO₂ = 65.798 amu

0.08868 moles x 65.798 = 5.83 g