Question

2B + 3H2 2BH3

How many grams of boron trihydride would be produced if 15.1 grams of hydrogen gas were reacted with an excess of boron?

Answer 1

x = 69.15 grams of BH3

Step-by-step explanation:

In order to find the answer for this question we need to know how many moles we have in 15.1 grams of Hydrogen gas, we will do that using its molar mass, which is 2.02 g/mol

2.02 g = 1 mol

15.1 g = x moles of H2

x = 7.5 moles of H2

Now we need to find the molar ratio between H2 and BH3, we can do that by looking at the reaction:

2 B + 3 H2 -> 2 BH3

So the reaction is showing us that for every 3 moles of H2 we will end up with 2 moles of BH3, which means we have a 3:2 molar ratio, but in our situation we actually have 7.5 moles of H2, now let's find out how many moles of BH3 we have

3 H2 = 2 BH3

7.5 H2 = x BH3

x = 5 moles of BH3

Now the last step will be finding the mass of BH3, we already have the number of moles and also the molar mass which is 13.83 g/mol, now we can calculate:

13.83 g = 1 mol

x grams = 5 mole

x = 69.15 grams of BH3