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How many milligrams of aluminum are present in 5.60 grams of aluminum sulfate?
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.
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.
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.
How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid With the balanced equation Al2O3 3H2SO4 -- Al2 (SO4)3 3H2O?
Al2O3 + 3H2SO4 = Al2 (SO4)3 + 3H2O We note that the molar ratios are 1:3::1:3 By Equivalence 0.89 : 2.67 ;; 0,89 :2.67 So moles of aluminium sulphate produced is 0.89 Using the equation moles = mass(g) / Mr Then mass = moles X Mr Refer to the Atomic Masses in the Periodic Table to find the Mr(Relative Molecular mass of Al2(SO4)3 Hence Al x 2 = 27 x 2 = 54 S x 3 = 32 x 3 = 96 O x (4 x 3 = 12) 16 x 12 = 192 54 + 96 + 192 = 342 ( Mr of Al2(SO4)3 ) Hence mass(g) = 0,89 X 342 = 304.38 g (This is the 100% theoretical yield. Experimentally it will be less , owing to losses in the filter paper , spillages etc., )
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).
How many milligrams of aluminum are present in 5.60 grams of aluminum sulfate?
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.
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.
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.
How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid With the balanced equation Al2O3 3H2SO4 -- Al2 (SO4)3 3H2O?
Al2O3 + 3H2SO4 = Al2 (SO4)3 + 3H2O We note that the molar ratios are 1:3::1:3 By Equivalence 0.89 : 2.67 ;; 0,89 :2.67 So moles of aluminium sulphate produced is 0.89 Using the equation moles = mass(g) / Mr Then mass = moles X Mr Refer to the Atomic Masses in the Periodic Table to find the Mr(Relative Molecular mass of Al2(SO4)3 Hence Al x 2 = 27 x 2 = 54 S x 3 = 32 x 3 = 96 O x (4 x 3 = 12) 16 x 12 = 192 54 + 96 + 192 = 342 ( Mr of Al2(SO4)3 ) Hence mass(g) = 0,89 X 342 = 304.38 g (This is the 100% theoretical yield. Experimentally it will be less , owing to losses in the filter paper , spillages etc., )
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).
How many moles of aluminum are produced from 33 grams of aluminum?
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.
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 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 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 could be produced from 34.0 grams of aluminum and 39.0 grams of chlorine gas?
To find the limiting reactant, we need to calculate the moles of each reactant. Then, use the stoichiometry of the balanced chemical equation to determine which reactant limits the amount of aluminum chloride that can be produced. Finally, calculate the mass of aluminum chloride produced based on the limiting reactant.
