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 H_{3}BO_{3} for every mole of Na_{2}B_{4}O_{7} produced. Using the given conversion factors:

150 tons of Na_{2}B_{4}O_{7} is 1.36x10^{8} grams.

The molar mass of Na_{2}B_{4}O_{7} is 201.2 g/mole, which means we have

6.77 x 10^{5} moles of Na_{2}B_{4}O_{7}

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

2.71 x 10^{6} moles of H_{3}BO_{3} (That's the metric term for a lotta H_{3}BO_{3})

At a molar mass of 61.8 g/mole for H_{3}BO_{3} , we can multiply the moles times molar mass to get grams:

1.67 x 10^{8} grams H_{3}BO_{3}

Bob