Determine the ratio of amount naoh reaction per unit amount fecl3. enter an integer that represents x mol naoh/1 mol fecl3. 1.50 mol naoh 0.50 mol fecl3

Here we have to calculate the amount of sodium hydroxide (NaOH) will react with ferric chloride (FeCl₃)

Thus 1 mole of FeCl₃ will react with 3 moles of NaOH. The integer will be 3. 0.739 g of NaOH per unit amount of FeCl₃.

Thus, 3 mol of NaOH/ 1 mol of FeCl3. 1.50 mol of NaOH per 0.50 mol of FeCl3.

The reaction between NaOH and FeCl₃ produces sodium chloride (NaCl) and ferric hydroxide [Fe (OH)₃]. The reaction is –  

3NaOH + FeCl₃ = 3 NaCl + Fe(OH)₃.

The molecular weight of NaOH and FeCl₃ are 39.99 g/mol and 162.2 g/mol. As per the reaction 1 mole of FeCl₃ will react with 3 moles of NaOH.

Thus 162.2 g of FeCl3 will react with (3×39.99) = 119.97 g of NaOH. So, = 0.739 g per unit FeCl₃.

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

i dont know

explanation: