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How many grams of aluminum is required to produce 25.0 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.
How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid?
To find the grams of aluminum sulfate produced, you would need to know the stoichiometry of the reaction between sulfuric acid and aluminum sulfate. Without that information, we cannot determine the exact amount of aluminum sulfate produced.
6.7 grams of aluminum sulfate to moles?
To convert grams of aluminum sulfate to moles, you first need to determine the molar mass of aluminum sulfate (Al2(SO4)3), which is approximately 342.15 g/mol. Then, divide the given mass by the molar mass to obtain the number of moles. In this case, 6.7 grams of aluminum sulfate is approximately 0.02 moles.
What is the mass of 0.25 mole of aluminum sulfate?
The molar mass of aluminum sulfate is 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be 85.54 grams (0.25 moles x 342.15 g/mol).
What is the molar mass of aluminum sulfate to the nearest gram?
The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342 grams per mole.
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?
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.
How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid?
To find the grams of aluminum sulfate produced, you would need to know the stoichiometry of the reaction between sulfuric acid and aluminum sulfate. Without that information, we cannot determine the exact amount of aluminum sulfate produced.
6.7 grams of aluminum sulfate to moles?
To convert grams of aluminum sulfate to moles, you first need to determine the molar mass of aluminum sulfate (Al2(SO4)3), which is approximately 342.15 g/mol. Then, divide the given mass by the molar mass to obtain the number of moles. In this case, 6.7 grams of aluminum sulfate is approximately 0.02 moles.
What is the mass of 0.25 moles of aluminum sulphate?
To find the mass of 0.25 moles of aluminum sulfate, you need to know the molar mass of aluminum sulfate. The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be around 85.54 grams.
What is the mass of 0.25 mole of aluminum sulfate?
The molar mass of aluminum sulfate is 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be 85.54 grams (0.25 moles x 342.15 g/mol).
What is the molar mass of aluminum sulfate to the nearest gram?
The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342 grams per mole.
What mass of nitrogen in grams is present in 148 grams of ammonium sulfate?
Ammonium sulfate contains 21% nitrogen by mass. To find the mass of nitrogen in 148 grams of ammonium sulfate, you would first calculate 21% of 148 grams, which equals 31.08 grams of nitrogen.
How many grams of copper ii sulfate are required to produce 0.48 mol of aluminum iii sulfate?
To find the amount of copper (II) sulfate needed to react with 0.48 mol of aluminum (III) sulfate, start by writing a balanced chemical equation for the reaction between the two salts. From the balanced equation, determine the molar ratio between copper (II) sulfate and aluminum (III) sulfate. Then, use this ratio to calculate the amount of copper (II) sulfate needed to produce 0.48 mol of aluminum (III) sulfate.
How many milligrams in 3.5 grams?
1 gram = 1,000 milligrams 2 grams = 2,000 milligrams 3 grams = 3,000 milligrams 3.5 grams = 3,500 milligrams
How many milligrams to 2.75 grams?
There are 1000 milligrams in a gram. Therefore 2.75 grams is 2750 milligrams.
How much caffeine is present in 10 grams of ground coffee?
There is about 100 milligrams of caffeine in 10 grams of ground coffee.
