Now consider the reaction when 45.0 g naoh have been added. what amount of naoh is this, and what amount of fecl3 can be consumed by it?

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain