how many grams are in a mole of soda cans? show work
There are approximately 25 empty 12-ounce aluminum cans that make up a pound.
To determine how many moles of aluminum are produced from 33 grams, divide the given mass by the molar mass of aluminum, which is approximately 26.98 g/mol. So, 33 g / 26.98 g/mol ≈ 1.22 moles of aluminum are produced.
In aluminum sulfate, the molar mass of aluminum is 27 g/mol. Calculate the amount of aluminum in 5.60 g of aluminum sulfate using the molar ratio between aluminum and aluminum sulfate (1:1). Therefore, there are 5.60 grams of aluminum in 5.60 grams of aluminum sulfate.
To find the amount of aluminum needed to produce aluminum sulfate, you need to consider the molar mass of aluminum sulfate and the ratio of aluminum in the compound. First, calculate the molar mass of aluminum sulfate (Al2(SO4)3). Then, find the ratio of aluminum in the compound (2 moles of Al in 1 mole of Al2(SO4)3). Finally, use this information to calculate the grams of aluminum needed to produce 25.0 grams of aluminum sulfate.
To find the amount of aluminum oxide that can be made, we need to determine the limiting reactant. The balanced chemical equation for the reaction is 4Al + 3O2 → 2Al2O3. The molar mass of aluminum oxide is 101.96 g/mol. After determining the limiting reactant and doing the stoichiometry calculation, we find that 100 grams of aluminum can produce 197 grams of aluminum oxide in this reaction.
1000 grams
Roughly 3,300 empty (obviously) 12 oz. aluminum cans.
1,99 grams of aluminum is equal to 0,0737 moles.
100
Your question needs to include the weight of each can before an answer to be formulated.
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
7
There are approximately 25 empty 12-ounce aluminum cans that make up a pound.
Rounded: 30 cans.
This is an experiment you may conduct for yourself. Just get a few dozen cans, and weigh them. Hint, 40 should be enough.
About one litre will do them.
To determine the grams of aluminum hydroxide obtained from 17.2 grams of aluminum sulfide, we need to consider the stoichiometry of the reaction between aluminum sulfide and water to form aluminum hydroxide. Given the balanced chemical equation, we can calculate the molar mass of aluminum hydroxide and use it to convert the mass of aluminum sulfide to grams of aluminum hydroxide formed.