calculate the moles of indicated product produced when 26.0g of each reactant is used.
A. 4Al+3O2---->2AlO3 (Al2O3
B. 3NO2+H2O--->2HNO3+NO (HNO3)
These are limiting reagent problems; I know that because amounts are given for BOTH reactants.
Convert 26 g Al to mols. mols = grams/molar mass.
Do the same for 26 g O2.
Using the coefficients in the balanced equation, convert mols Al to mols Al2O3.
Do the same with mols O2 to mols Al2O3.
It is quite likely that the two values for mols Al2O3 will not be the same which means one of them is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Then convert the smaller value to grams. g = mols Al2O3 x molar mass Al2O3.
To calculate the moles of the indicated products produced, you need to follow these steps:
Step 1: Determine the molar mass of the reactants and products involved in the reaction. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). You can find the atomic masses of the elements on the periodic table.
Step 2: Calculate the number of moles of each reactant used. This can be done by dividing the given mass of the reactant by its molar mass.
Step 3: Determine the stoichiometric ratio between the reactant and the product. This is obtained from the balanced chemical equation and represents the ratio of moles of reactant to product.
Step 4: Calculate the moles of the indicated product produced by multiplying the moles of the reactant used by the stoichiometric ratio.
Now let's apply these steps to the given reactions:
A. 4Al + 3O2 → 2Al2O3
Step 1: The molar mass of Al is 26.98 g/mol, and the molar mass of Al2O3 is 101.96 g/mol.
Step 2: Moles of Al = Mass of Al / Molar mass of Al
= 26.0 g / 26.98 g/mol
≈ 0.963 moles of Al
Step 3: The stoichiometric ratio between Al and Al2O3 is 4:2, or simplified to 2:1. This means that for every 2 moles of Al, 1 mole of Al2O3 is produced.
Step 4: Moles of Al2O3 = Moles of Al * Stoichiometric ratio
= 0.963 moles * (1 mole Al2O3 / 2 moles Al)
= 0.4815 moles Al2O3
Therefore, approximately 0.4815 moles of Al2O3 are produced.
B. 3NO2 + H2O → 2HNO3 + NO
Step 1: The molar mass of NO2 is 46.0 g/mol, and the molar mass of HNO3 is 63.01 g/mol.
Step 2: Moles of NO2 = Mass of NO2 / Molar mass of NO2
= 26.0 g / 46.0 g/mol
≈ 0.565 moles of NO2
Step 3: The stoichiometric ratio between NO2 and HNO3 is 3:2. This means that for every 3 moles of NO2, 2 moles of HNO3 are produced.
Step 4: Moles of HNO3 = Moles of NO2 * Stoichiometric ratio
= 0.565 moles * (2 moles HNO3 / 3 moles NO2)
≈ 0.3767 moles HNO3
Therefore, approximately 0.3767 moles of HNO3 are produced.