Question

What is the experimental yield (in grams) of the solid product when the percent yield is 83.5% with 11.23 g of iron(II) nitrate reacting in solution with excess sodium phosphate?

Fe(NO₃)₂(aq) + Na₃PO₄(aq) --> Fe₃(PO₄)₂(s) + NaNO₃(aq) [unbalanced]

What is the percent yield of the solid product when 10.95 g of barium nitrate reacts in solution with excess sodium phosphate and 4.185 g of the precipitate is experimentally obtained?

Ba(NO₃)₂(aq) + Na₃PO₄(aq) --> Ba₃(PO₄)₂(s) + NaNO₃(aq) [unbalanced]

What is the percent yield of the solid product when 18.62 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 4.876 g of the precipitate is experimentally obtained?

Fe(NO₃)₃(aq) + Na₃PO₄(aq) --> FePO₄(s) + NaNO₃(aq) [unbalanced]

Answer 1

1) Given, the reaction,

Fe(NO₃)₂(aq) + Na₃PO₄(aq)   Fe₃(PO₄)₂(s) + NaNO₃(aq)

Balancing the above reaction,

3Fe(NO₃)₂(aq) + 2Na₃PO₄(aq)   Fe₃(PO₄)₂(s) + 6NaNO₃(aq)

Also given,

Mass of iron(II) nitrate = 11.23 g

Percent yield = 83.5 %

Firstly calculating the number of moles of iron(II) nitrate from the given mass,

= 11.23 g iron(II) nitrate x (1 mol /179.855 g)

= 0.06244 mol of iron(II) nitrate

Now, Using the calculated moles and mole ratio from the balanced chemical reaction, calculating the number of moles of solid product forming,

= 0.06244 mol of iron(II) nitrate x ( 1 mol of Fe₃(PO₄)₂ / 3 mol of  iron(II) nitrate)

= 0.02081 mol of Fe₃(PO₄)₂

Converting number of moles of Fe₃(PO₄)₂ to grams,

= 0.02081 mol of Fe₃(PO₄)₂ x (357.48 g /1 mol)

= 7.44 g of Fe₃(PO₄)₂

Theoretical yield = 7.44 g

Also given, Percent yield = 83.5%

We know,

Percent yield = (Experimental yield / Theoretical yield) x 100

83.5 = (Experimental yield / 7.44) x 100

Experimental yield = 6.21 g

2) Given, the reaction,

Ba(NO₃)₂(aq) + Na₃PO₄(aq)   Ba₃(PO₄)₂(s) + NaNO₃(aq)

Balancing the above reaction,

3Ba(NO₃)₂(aq) + 2Na₃PO₄(aq)   Ba₃(PO₄)₂(s) + 6NaNO₃(aq)

Also given,

Mass of barium nitrate = 10.95 g

Experimental yield = 4.185 g

Firstly calculating the number of moles of barium nitrate from the given mass,

= 10.95 g barium nitrate x (1 mol /261.337 g)

= 0.0419 mol of barium nitrate

Now, Using the calculated moles and mole ratio from the balanced chemical reaction, calculating the number of moles of solid product forming,

= 0.0419 mol of barium nitrate x ( 1 mol of Ba₃(PO₄)₂ / 3 mol of barium nitrate)

= 0.01396 mol of Ba₃(PO₄)₂

Converting number of moles of Fe₃(PO₄)₂ to grams,

= 0.01396 mol of Ba₃(PO₄)₂ x (601.92 g /1 mol)

= 8.41 g of Ba₃(PO₄)₂

Theoretical yield = 8.41 g

Also given, Experimental yield = 4.185 g

We know,

Percent yield = (Experimental yield / Theoretical yield) x 100

Percent yield = (4.185 g / 8.41 g ) x 100

Percent yield = 49.8%

3)  Given, the reaction,

Fe(NO₃)₃(aq) + Na₃PO₄(aq)   FePO₄(s) + NaNO₃(aq)

Balancing the above reaction,

Fe(NO₃)₃(aq) + Na₃PO₄(aq)   FePO₄(s) + 3NaNO₃(aq)

Also given,

Mass of iron(III) nitrate = 18.62 g

Experimental yield = 4.876 g

Firstly calculating the number of moles of iron(III) nitrate from the given mass,

= 18.62 g iron(III) nitrate x (1 mol /241.86 g)

= 0.07698 mol of iron(III) nitrate

Now, Using the calculated moles and mole ratio from the balanced chemical reaction, calculating the number of moles of solid product forming,

= 0.07698 mol of iron(III) nitrate x ( 1 mol of FePO₄ / 1 mol of  iron(III) nitrate)

= 0.07699 mol of FePO₄

Converting number of moles of FePO₄ to grams,

= 0.07699 mol of FePO₄ x (150.82 g /1 mol)

= 11.6 g of FePO₄

Theoretical yield = 11.6 g

Also given, Experimental yield = 4.876 g

We know,

Percent yield = (Experimental yield / Theoretical yield) x 100

Percent yield = (4.876 g / 11.6 g) x 100

Percent yield = 42.0 %