Answer:
Grams of AlCl₃ ~ 80 grams (1 sig. fig.)
Explanation:
Write and balance given equation.
Coefficients should be in smallest whole number ratios. => called ‘The Standard Equation’ and is assumed to be at 0˚C & 1atm. pressure.
3NaOH + AlCl₃ => Al(OH)₃ + 3NaCI
Write below compound what’s given and what’s needed, i.e., to be calculated.
Rxn: 3NaOH + AlCl₃ => Al(OH)₃ + 3NaCI
Given: --------- (?g) -------- 100.0g
Rxn Moles: 3 moles 1 mole 1 mole 3 moles
Calc’d Moles: --------- (? Moles) -------- 100.0g/58g∙molˉ¹
= 1.724 mol NaCl
Calculate moles AlCl₃ needed using mole ratios:
(? moles AlCl₃/1 mole AlCl₃) = (1.724 moles NaCl/3 moles NaCl)
=> ? moles AlCl₃ = (1.724 moles NaCl x 1 mole AlCl₃) / (3 moles NaCl)
=> ? moles AlCl₃ = (1.724 x 1 / 3) mole AlCl₃ = 0.575 mole AlCl₃ needed
Convert moles AlCl₃ needed to grams AlCl₃ needed:
Grams AlCl₃ needed = calculated moles AlCl₃ x formula wt. AlCl₃
= 0.575 mole AlCl₃ x 133.34 grams AlCl₃/mole AlCl₃
= 76.6705 grams AlCl₃ needed (calculated answer)
Express answer in appropriate number of sig. figs.
- Number of sig. figs. for answer = the given measured data having the least number of sig. figs.
- In this case, grams of NaCl = 100.0 grams (measured data) contains only 1 sig. fig. (Do not use conversion factors or universal constants)
- Therefore the final answer => Grams of AlCl₃ ~ 80 grams (1 sig. fig.)
