AlPO4
6.5 moles AlPO4 (121.95 grams/1 mole AlPO4)
= 792.675 grams of AlPO4 ( you do sigi figis )
To convert moles to grams, you need to use the molar mass of the substance. The molar mass of aluminum phosphate is 122.94 g/mol. Therefore, for 5.5 moles of aluminum phosphate, you would have 5.5 moles x 122.94 g/mol = 676.17 grams of aluminum phosphate.
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.
The molar mass of aluminum is approximately 27.0 g/mol. To find the mass of 3.0 moles of aluminum, you can multiply the number of moles by the molar mass: 3.0 moles * 27.0 g/mol = 81.0 grams of aluminum.
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.
Given the balanced chemical equation: 4Al + 3O2 → 2Al2O3, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. In this case, 18.32 grams of aluminum is equivalent to 0.684 moles. Using stoichiometry, we find that this would produce 0.456 grams of aluminum oxide.
To convert moles to grams, you need to use the molar mass of the substance. The molar mass of aluminum phosphate is 122.94 g/mol. Therefore, for 5.5 moles of aluminum phosphate, you would have 5.5 moles x 122.94 g/mol = 676.17 grams of aluminum phosphate.
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 ---------------------------------
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
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.
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
To find the mass of calcium phosphate (Ca₃(PO₄)₂) in grams for 0.658 moles, first calculate its molar mass. The molar mass of calcium phosphate is approximately 310.18 g/mol. Multiply the number of moles by the molar mass: 0.658 moles × 310.18 g/mol ≈ 204.4 grams. Thus, there are about 204.4 grams of calcium phosphate in 0.658 moles.
The molar mass of aluminum is approximately 27.0 g/mol. To find the mass of 3.0 moles of aluminum, you can multiply the number of moles by the molar mass: 3.0 moles * 27.0 g/mol = 81.0 grams of aluminum.
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.
Given the balanced chemical equation: 4Al + 3O2 → 2Al2O3, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. In this case, 18.32 grams of aluminum is equivalent to 0.684 moles. Using stoichiometry, we find that this would produce 0.456 grams of aluminum oxide.
For this you need the atomic mass of Al. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.91 moles Al × (27.0 grams) = 78.6 grams Al
The balanced equation for this reaction is: 2K3PO4 + 3Al(NO3)3 -> 6KNO3 + AlPO4. This indicates that 2 moles of potassium phosphate react with 2 moles of aluminum nitrate to produce 6 moles of potassium nitrate.