Question

if 23 grams of iron (II) reacts with 41 grams of sodium phosphate, how much sodium chloride can be formed

Answer 1

The 45.1 grams of sodium chloride can be formed. To find how much sodium chloride can be formed when 23 grams of iron (II) reacts with 41 grams of sodium phosphate, the balanced chemical equation and stoichiometry are used to calculate the mass of sodium chloride. The calculation involves converting the given mass of iron to moles, using the stoichiometric ratio to find the moles of sodium phosphate, and then converting the moles of sodium phosphate to mass.

Step-by-step explanation:

To determine how much sodium chloride can be formed when 23 grams of iron (II) reacts with 41 grams of sodium phosphate, we need to balance the chemical equation first.

The balanced chemical equation for the reaction between iron (II) and phosphate is:

3 Fe + 2 Na3PO4 → Fe3(PO4)2 + 6 Na

From the balanced equation, we can see that the stoichiometric ratio between Fe and Na3PO4 is 3:2.

This means that for every 3 moles of Fe, we need 2 moles of Na3PO4 to react.

To solve the problem, we can follow these steps:

1. Convert the given mass of Fe to moles using the molar mass of Fe.
2. Convert the moles of Fe to moles of Na3PO4 using the stoichiometric ratio.
3. Convert the moles of Na3PO4 to mass using the molar mass of Na3PO4.

Let's calculate:

Molar mass of Fe = 55.85 g/mol

Molar mass of Na3PO4 = 163.94 g/mol

Step 1:

Moles of Fe = mass of Fe / molar mass of Fe

= 23 g / 55.85 g/mol

= 0.412 mol

Step 2:

Moles of Na3PO4 = (moles of Fe) x (2 moles of Na3PO4 / 3 moles of Fe)

= (0.412 mol) x (2/3)

= 0.275 mol

Step 3:

Mass of Na3PO4 = moles of Na3PO4 x molar mass of Na3PO4

= (0.275 mol) x (163.94 g/mol)

= 45.1 g

Therefore, 45.1 grams of sodium chloride can be formed.