If I receive links for an answer you will be reported. For me this is due today at 11:59 and I really need help please. I asked this question before but received no answer. NaOH + AlCl3 --> Al(OH)3 + NaCI.
How many grams of AlCl3 are needed to produce 100.0 g NaCl?

76.1g

Explanation:

The balanced equation for this reaction is as follows:

3NaOH + AlCl3 --> Al(OH)3 + 3NaCI

From the above equation, it can be observed that 1 mole of AlCl3 is needed to produce 3 moles of NaCl.

Using the formula below, we convert the mass of NaCl to moles;

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

mole = 100/58.5

mole = 1.71mol of NaCl

Since 1 mole of AlCl3 is needed to produce 3 moles of NaCl.

Then 1.71 moles of NaCl will be produced by 1.71/3 = 0.57mol of AlCl3

Molar mass of AlCl3 = 27 + 35.5(3)

= 27 + 106.5

= 133.5g/mol

Mole = mass/molar mass

Mass = molar mass × mole

Mass = 133.5 × 0.57

Mass = 76.1g