Question

If you have 1.20g of Al, how many grams of potassium aluminum sulfate dodecahydrate should you...

If you have 1.20g of Al, how many grams of potassium aluminum sulfate dodecahydrate should you expect to produce if you have no other limiting reactants?

Homework Answers

Answer #1

we know that

moles = mass /molar mass

molar mass of Al = 27 g/mol

given

mass of Al = 1.2 g

so

moles of Al = 1.2 / 27

moles of Al = 0.044444

now

we know that

potassium aluminum sulfate dodecahydrate is KAl(S04)2 . 12 H20

we can see that

1 mole of Al can given 1 mole of KAl(S04)2 . 12 H20

so

moles of KAl(SO4)2 . 12 H20 = moles of Al = 0.04444

now

mass = moles x molar mass

molar mass of KAl(SO4)2 = 474.3884 g/mol

so

mass of KAl(S04)2 produced = 0.0444 x 474.3884 = 21.084

so

21.084 grams of potassium aluminum sulfate dodecahydrate can be obtained

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
According to the following reactions: A) how many grams of potassium sulfate are required for the...
According to the following reactions: A) how many grams of potassium sulfate are required for the complete reaction of 22.0 grams of barium chloride? barium chloride(aq) + potassium sulfate(aq) --> barium sulfate(s) + potassium chloride(aq) B) how many grams of hydrogen peroxide (H2O2) are needed to form 32.8 grams of oxygen gas? hydrogen peroxide (H2O2)(aq) --> water(ℓ) + oxygen(g)
The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate....
The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate. The instructions for the potassium aluminum sulfate preparation give a target of between 0.025 and 0.050 moles of the product. The instructions also specify to use an excess of potassium hydroxide but not more than 2 times the amount required by the balanced equation. Calculate the mass in grams of aluminum needed to prepare the maximum number of moles of product as noted above....
The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate....
The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate. The instructions for the potassium aluminum sulfate preparation give a target of between 0.025 and 0.050 moles of the product. The instructions also specify to use an excess of potassium hydroxide but not more than 2 times the amount required by the balanced equation. Calculate the mass in grams of aluminum needed to prepare the maximum number of moles of product as noted above....
If you start with 1.115g of aluminum, how many grams of alum should be obtained?
If you start with 1.115g of aluminum, how many grams of alum should be obtained?
Write the balanced reaction for the mixing of aqueous calcium nitrate and aqueous aluminum sulfate which...
Write the balanced reaction for the mixing of aqueous calcium nitrate and aqueous aluminum sulfate which produces solid calcium sulfate and aqueous aluminum nitrate. How many grams of aluminum sulfate would be needed to produce 1.5 grams of aluminum nitrate?
how many protons are in 891mg of aluminum sulfate?
how many protons are in 891mg of aluminum sulfate?
How many grams of solid potassium cyanide should be added to 1.00 L of a 0.239...
How many grams of solid potassium cyanide should be added to 1.00 L of a 0.239 M hydrocyanic acid solution to prepare a buffer with a pH of 10.174 ? grams potassium cyanide=
1. How many GRAMS of manganese(II) are present in 1.53 grams of manganese(II) sulfate ?   grams...
1. How many GRAMS of manganese(II) are present in 1.53 grams of manganese(II) sulfate ?   grams manganese(II). 2. How many GRAMS of manganese(II) sulfate can be made from 2.46 grams of manganese(II) ?   grams manganese(II) sulfate.
Elemental (metallic) aluminum (Al) is reacted with hydrochloric acid (HCl) to yield aluminum chloride and hydrogen...
Elemental (metallic) aluminum (Al) is reacted with hydrochloric acid (HCl) to yield aluminum chloride and hydrogen gas. How many grams of metallic aluminum would be required to produce 1 gram of pure hydrogen gas? REMEMBER to balance the reaction and show all steps.
1. If actual amount of alum obtained from 2.535 grams of aluminum sulfate is 3.5079 g...
1. If actual amount of alum obtained from 2.535 grams of aluminum sulfate is 3.5079 g , what is the percent yield of alum? 2. In a combination reaction, if 1.34794 moles of magnesium is heated with 0.2638 moles of nitrogen, to form magnesium nitride, how many grams of magnesium nitride will be produced? If actual amount of alum obtained from 2.535 grams of aluminum sulfate is 3.5079 g , what is the percent yield of alum?