Consider the following reaction:
2 B5H9 + 12 O2 => 5 B2O3 + 9 H2O
How many grams of B2O3 are made by the reaction of 233.8 g of O2
The steps are exactly the same as your previous question about H2O. You would do well to remember those steps. They work for all stoichiometry problems.
To find the grams of B2O3 produced by the reaction of 233.8 g of O2, you can use stoichiometry. Stoichiometry is a branch of chemistry that deals with the calculation of reactants and products in chemical reactions.
In this reaction, the stoichiometric ratio between O2 and B2O3 is 12:5. This means that for every 12 moles of O2, 5 moles of B2O3 are produced.
To solve this problem, you will follow these steps:
1. Convert the given mass of O2 (233.8 g) to moles. To do this, divide the mass by the molar mass of O2.
The molar mass of O2 is 16 g/mol (since oxygen has a molar mass of 16 g/mol and there are two oxygen atoms in O2). Therefore, the number of moles of O2 is 233.8 g / 16 g/mol = 14.61 mol.
2. Use the stoichiometric ratio to determine the number of moles of B2O3 produced from the moles of O2.
Since the stoichiometric ratio is 12:5 (O2:B2O3), divide the number of moles of O2 by the ratio:
14.61 mol O2 * (5 mol B2O3 / 12 mol O2) = 6.05125 mol B2O3.
3. Convert the moles of B2O3 to grams. To do this, multiply the number of moles by the molar mass of B2O3.
The molar mass of B2O3 is 69.62 g/mol (boron has a molar mass of 10.81 g/mol, and there are two boron atoms in B2O3). Therefore, the mass of B2O3 is:
6.05125 mol * 69.62 g/mol = 421.53 g.
Therefore, the reaction of 233.8 g of O2 will produce 421.53 g of B2O3.