I really need help please. NaOH + AlCl3 --> Al(OH)3 + NaCI.
How many grams of AlCl3 are needed to produce 100.0 g NaCl?

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.

1. they vary with a higher number because boiling is way hotter than melting

the answer is a and have a great day