Question

AICI HO not balanced) 14. Using the equation: HCI + Al(OH). How many grams of AICI, will you make if you start with 234.03 grams of Allow a) What do you have and what do you need b) Write the unbalanced equation c) Balance the equation d) From 234,03 grams, how many moles of Al(OH), do you have? (Convert grams to moles) Molecule: fone molecule) Atomic Breakdown: counting atoms skill) # Moles = 1 Atomic weight of element a tom of element in 1 molecule Atomic weight of element in molecule mole B/mole Number moles Mol wgt e) Ratio the balanced equation with the newly calculated number of moles f) From this newly generated equation, how many moles of AICI, do you have? s) How many grams of AlCl3 did you make? (Convert moles to grams) Molecule: (one molecule) Atomic Breakdown: (counting atoms skill) # Moles = 1 Atomic weight of element #atom of element in 1 Element g/mole molecule Atomic weight of element in molecule g/mole sum Number moles Mol wgt

Answer 1

solution (A) (a) we have unbalanced equation, - Mass of alcorda = 234'03 g We need formula to calculate number of moles Atomic weight of elements : Hill, Ad, do. Atomic weight of H = 18 lmal ce= 35.5 g/mol o= 16 g/mol Al = 27 g/mol Number of moles = mass molar mass (6) Unbalanced equation : B He + Al(OH)3 - AlCl3 + H₂O (c) since 3 ce atoms on right hand side so to balance it multiply by 3x Hce => 3 HCl + Al(OH)3 - Alllz + H2O Now 3,0 atoms on left hand side so to balance it multiply by 3 X H₂O =) 3HCl + Allow)3 Alcl3 + 3H2O = Balanced equation ; 3HCl + Aul (OH)3 = Alcl3+3HD (d) in 234.03 qram of AllOH)3 we need to - calculate its number of moles. Now molare mass of Allow), = 27 +1683 +183 1 = 78 g/mol number of moles of Al(OH)3 = 2.34.03g - 78 g/mol = 13 molds