1. Stoichiometry Practice Problem Nick Maskrey

2. Problem If 5.26 g of Lithium Sulfate react with Sodium Phosphate complete the stoichiometry. If 5.26 g of Lithium Sulfate react with Sodium Phosphate complete the stoichiometry.

3. Complete and Balance the Equation 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 )

4. Split Each Chemical Into Separate Columns 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 )

5. Write the given measurement into the correct column 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g.

6. Convert the given into moles 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles

7. Get moles for all other columns by putting the moles of the given times (x) a fraction in each 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles.0478 x /

8. The numerator is the coefficient of the column 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles.0478 x 2/.0478 x 3/

9. The denominator is the coefficient of the given 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles.0478 x 2/3.0478 x 3/3

10. Do all of the math to get the moles 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles.0478 x 2/3.0319 moles.0478 x 2/3.0319 moles.0478 x 3/3.0478 moles

11. Convert All to Grams 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles.0478 x 2/3.0319 moles.0319 m x 163.939g 1 1 mole 3(22.989) + 30.973 + 4(15.994) 163.938 g 5.23 g.0478 x 2/3.0319 moles.0319 m x 115.994g 1 1 mole 3(6.941) + 30.973 + 4(15.9994) 115.794 g 3.69 g.0478 x 3/3.0478 moles.0478m x 142.041g 1 1 mole 2(22.989) + 32.065 +4(15.9994) 142.041 g 6.79 g

12. Verify the law of conservation of mass 3Li 2 (SO4) + 2Na 3 (PO 4 ) ---> 2Li 3 (PO 4 ) + 3Na 2 (SO 4 ) 5.26 g. 5.26g x 1 mole 1 109.945 g 2(6.941) + 32.065 + 4(15.9994) 109.945 g.0478 moles 5.26g + 5.23g 10.49g.0478 x 2/3.0319 moles.0319 m x 163.939g 1 1 mole 3(22.989) + 30.973 + 4(15.994) 163.938 g 5.23 g.0478 x 2/3.0319 moles.0319 m x 115.994g 1 1 mole 3(6.941) + 30.973 + 4(15.9994) 115.794 g 3.69 g 3.69g + 6.79 g 10.48g.0478 x 3/3.0478 moles.0478m x 142.041g 1 1 mole 2(22.989) + 32.065 +4(15.9994) 142.041 g 6.79 g

13. Answer The reactant mass adds up to 10.49 g and the products mass adds up to 10.48 g. The third significant digit in each is 4 and within two of each other so the problem is correct. The reactant mass adds up to 10.49 g and the products mass adds up to 10.48 g. The third significant digit in each is 4 and within two of each other so the problem is correct.