so I need help with stoichiometry#2 is the lesson we are on if anyone could please explain. the problem is down below

Question

Sodium tetraborate is produced according to the following reaction:

4H₃BO₃ + 2NaOH → 7H₂O + Na₂B₄O₇

How many grams boric acid are needed to make 150 tons of sodium tetraborate? 

(Remember that 454 g = 1 lb and 2000 lbs = 1 ton)

Answer 1

Hello, Luna,

grams.

There are some large numbers here, since we are asked to answer in grams when the output is expressed in tons. The approach I'm taking is to calculate the number of moles 150 tons represents and then multiply that by 4 since we know we need 4 moles of H₃BO₃ for every mole of Na₂B₄O₇ produced. Using the given conversion factors:

150 tons of Na₂B₄O₇ is 1.36x10⁸ grams.

The molar mass of Na₂B₄O₇ is 201.2 g/mole, which means we have

6.77 x 10⁵ moles of Na₂B₄O₇

From the balanced equation, we need 4 times this number of moles:

2.71 x 10⁶ moles of H₃BO₃ (That's the metric term for a lotta H₃BO₃)

At a molar mass of 61.8 g/mole for H₃BO₃ , we can multiply the moles times molar mass to get grams:

1.67 x 10⁸ grams H₃BO₃

Bob