It is as 'big' as 0.01 mole Al (not the 'rodent' mole, that's much bigger than 0.27 g!)
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
266,86 g aluminium chloride are obtained.
Thast is one mole. 26.98 grams/per mole is the mass.
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
9.4 grams
How much is 5 grams of aluminum
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.
1,99 grams of aluminum is equal to 0,0737 moles.
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
The molar mass of anhydrous aluminum chloride is 133,34 grams.
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.
To find the grams of aluminum hydroxide from 15.7 grams of aluminum sulfide, you first need to balance the chemical equation. The balanced equation is 2Al2S3 + 6H2O -> 4Al(OH)3 + 3H2S. Next, calculate the molar mass of aluminum sulfide (Al2S3) and aluminum hydroxide (Al(OH)3), then use the stoichiometry from the balanced equation to find the grams of aluminum hydroxide produced.
To determine the grams of aluminum oxide formed, we need to consider the balanced chemical equation for the reaction between aluminum and oxygen. The molar ratio between aluminum and aluminum oxide is 4:2. So, first calculate the moles of aluminum in 1020g, then use this to find the moles of aluminum oxide produced, and finally convert moles of aluminum oxide to grams.
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