What is the theoretical yield of aluminum oxide if 2.40 mol of aluminum metal is exposed to 2.10 mol of oxygen?<span><span>Ask for details </span><span>Follow </span>Report</span><span> by <span>Sapperd2248 </span><span>1 minute ago
1879</span></span>
TLDR: 6.53x10^5 g NH4ClO4
The stoichiometric coefficients (the numbers in front of the reactants and products) show that Aluminum and Ammonium Perchlorate are consumed at the exact same rate throughout the reaction: 3 parts of one to 3 parts of another.
1.5x10^5 grams of Aluminum, considering that the formula weight of Aluminum is 26.98 g/mol, is equal to 5,559.7 moles of Aluminum. This means that 5,559.7 moles of Ammonium Perchlorate are required to run the reaction to completion.
The formula weight of Ammonium Perchlorate is 117.49 grams a mole, and multiplying it by 5,559.7 moles to react to completion means that 6.53x10^5 grams of Ammonium Perchlorate is required for the reaction.
<span>4.3065 g
First, lookup the atomic weights of all the elements involved.
Atomic weight of Calcium = 40.078
Atomic weight of Carbon = 12.0107
Atomic weight of Hydrogen = 1.00794
Atomic weight of Oxygen = 15.999
Atomic weight of Sulfur = 32.065
Now calculate the molar masses of the reactants and product
Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999
= 98.07688 g/mol
Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999
= 100.0857 g/mol
Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999
= 136.139 g/mol
The balanced reaction for H2SO4 with CaCO3 is
CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2
So it takes 1 mole each of CaCO3 and H2SO4 to produce 1 mole of CaSO4. Let's see how many moles of CaCO3 and H2SO4 we have.
CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol
H2SO4: 3.2900g / 98.07688 g/mol = 0.033545113 mol
We have a slight excess of H2SO4, so the amount of CaCO3 is the limiting reactant and we should have 0.031632891 moles of product. To determine its mass, multiply the number of moles by the molar mass computed earlier.
0.031632891 mol * 136.139 g/mol = 4.306470148 g
Since we have 5 significant figures in our data, round the result to 5 figures, giving 4.3065 g</span>
6 electrons... 's' can hold 2..... 'd' can hold 10 and 'f' can hold 14