I am a parent trying to help with my childs homework. This is the question that has us stuck. Can you assist us?
Boron trioxide B203 , reacts with waterto produce boric acid, B(OH)3.
B2O3(s) + 3H2o (l) -> 2B(OH)3(s)
Ricardo adds 27.2 g of B2O3 to an excess of h2O, and the reationgoes to completion. How many moles ofB(OH)3 does the reation produce?
What is the formula I should be using to work out this problem?
This is a regular stoichiometry problem. All of these are worked with the following steps.
1. Write and balance the equation. You have that.
2. Convert grams of what you have (in this case B2O3) to mols. mols = grams B2O3/molar mass B2O3 = ? I'm estimating mols = approx 0.4 but you need to do the math to get an accurate answer.
3. Using the coefficients in the balanced equation, convert mols of what you have (B2O3) into mols what you want (in this case B(OH)3). That is 0.4 x (2 mol B(OH)3/1 mol B2O3) = 0.4 x 2/1 = approx 0.8 mol. Again, an estimate.
4. That's the problem and the answer is approx 0.8. MOST problems of this type don't ask for mols but ask for grams. So step 4 converts to grams in case your next problem is that kind.
g B(OH)3 = mols B(OH)3 x molar mass B(OH)3.
To solve this problem, you need to use the concept of stoichiometry, which is the relationship between the moles of reactants and products in a chemical reaction. In this case, you want to determine the number of moles of B(OH)3 produced.
First, you should convert the given mass of B2O3 (27.2 g) to moles. To do this, you need the molar mass of B2O3, which is calculated by finding the atomic masses of boron (B) and oxygen (O) from the periodic table and multiplying by the respective subscripts. The molar mass of B2O3 is approximately 69.62 g/mol.
To convert grams to moles, use the formula:
moles = mass (g) / molar mass (g/mol)
So, moles of B2O3 = 27.2 g / 69.62 g/mol ≈ 0.3911 mol
According to the balanced equation, the reaction stoichiometry shows that 1 mole of B2O3 reacts to produce 2 moles of B(OH)3. Therefore, you can conclude that 0.3911 mol of B2O3 will produce (0.3911 mol) x (2 mol B(OH)3 / 1 mol B2O3) ≈ 0.7822 mol of B(OH)3.
Hence, the reaction will produce approximately 0.7822 moles of B(OH)3.