AlPO4
6.5 moles AlPO4 (121.95 grams/1 mole AlPO4)
= 792.675 grams of AlPO4 ( you do sigi figis )
671
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
Because the atomic mass of aluminum is 26.99, this means that there are 26.99 grams of aluminum in one mole. For 3 moles multiply 26.99 by 3 = 80.97grams.
For this you need the atomic (molecular) mass of Al2O3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. Al2O3= 102 grams408 grams Al / (102 grams) = 4.00 moles Al
Since you are using 0.28 moles of AlCl3, you need 0.84 moles of Br, which is 67.24 grams.
671
1,99 grams of aluminum is equal to 0,0737 moles.
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
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).
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
Because the atomic mass of aluminum is 26.99, this means that there are 26.99 grams of aluminum in one mole. For 3 moles multiply 26.99 by 3 = 80.97grams.
there are two moles produced in potassium nitrate.
The formular you need is M = n/m (molar mass = amount of substance/mass), or n = m/MAluminium has a molar mass of 26.982 g/mol. The given mass is 3 grams. Therefore:n(Al) = 3g / (26.982g/mol) = 0.11mol
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
735 g of Ca3(PO4)2 are obtained.
all you have to do to fine the moles of any element is divide the given grams by the molar mass (which you can find on a periodic table of elements) Likewise, if you need to find the grams, just multiply the number of moles by molar mass. moles = grams/molar mass grams = moles x molar mass your equation should look like this: moles = 89.0 / 17.0 moles = .471