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How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid?
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How many grams of aluminum is required to produce 25.0 grams of aluminum sulfate?
2
How many grams of sodium sulfate will be formed if you start with 12.5 grams of sodium hydroxide and you have an excess of sulfuric acid?
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
How many milligrams of aluminum are present in 5.60 grams of aluminum sulfate?
The answer is 1 527 mg aluminium.
How many moles of aluminum are produced from 33 grams of aluminum?
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
400.0 grams of ammonia react with 56.0 moles of sulfuric acid to produce ammonium sulfate?
44.1-44.5 -Apex
How many moles are 4.12 grams aluminum sulfate?
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
How many grams of aluminum is required to produce 25.0 grams of aluminum sulfate?
2
How many grams of sodium sulfate will be formed if you start with 12.5 grams of sodium hydroxide and you have an excess of sulfuric acid?
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
How many milligrams of aluminum are present in 5.60 grams of aluminum sulfate?
The answer is 1 527 mg aluminium.
How many moles of aluminum are produced from 33 grams of aluminum?
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
The molar mass of aluminum sulfate?
Aluminium Sulphate is Al2(SO4)3 and its mass is27x2 + 96x3 = 342g/mole
400.0 grams of ammonia react with 56.0 moles of sulfuric acid to produce ammonium sulfate?
44.1-44.5 -Apex
How many grams of aluminum chloride are produced when 54 grams of aluminum react with chlorine gas?
266,86 g aluminium chloride are obtained.
How many grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloride acid?
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
The formula for sulfuric acid is H2SO4 A mole of sulfuric acid would have a mass of grams?
It would be 98 grams.
What is the osmolarity of 10.00 grams of aluminum sulfate dissolved in 200ml of aqueous solution?
To find this answer it is necessary to first find the chemical formula of aluminum sulfate utilizing the valence charges from the periodic table. Aluminum Sulfate is a molecular formed by the ionic bonding of aluminum to to the polyatomic ion of sulfate Al = +3 SO4 is a polyatomic ion with a charge of -2 Therefore after balancing the equation, you get the chemical formula Al2(SO4)3 Next, in order to find osmolarity, you must first find molarity. This is done by converting grams to moles. Moles are calculated in grams/Liter and is equal to the molecular weight. What is the molecular weight of aluminum sulfate. Looking at the periodic table, you sum the atomic weights of each element, shown in AMUs. Al = 27 (x2) = 54 S = 32 (x3) =96 O = 16 (x12) = 192 We add these up to determine the molecular weight of the substance, which can be used to determine the molarity of this solution. MW = 342g, which means that one mole of aluminum sulfate is 342 grams per liter of water. We can now use the information presented in the above problem to determine molarity 10 grams dissolved in 200 ml of water. Since measurements are in grams per liter, we can assume that if 10 grams were dissolved in 200 ml of water, then 50 grams will dissolve in 1000 ml, or 50 grams per liter (multiply both factors by five). We must use our grams of solute to convert grams into our molarity 50 g x 1 mole/342 grams of aluminum sulfate to determine the molarity. = .15 M At this point, to determine osmolarity, you must be aware that osmolarity is the measure of PARTICLES within a solution per liter. Realizing that aluminum sulfate is held together by an ionic bond, it will freely dissolve in water and produce Al ions and SO4 ions. So we sum the particles of each: 2 Al ions per molecule 3 SO4 ions per molecule Therefore, multiplying the molarity of the aluminum sulfate solution times 5 (total 5 particles from each molecule) will give you the osmolarity. = 0.73 Osm per liter of aqueous solution.
The formula for sulfuric acid is HËÃSOËÅ A mole of sulfuric acid would have a mass of grams?
It would have a mass of 98 grams.
