Percent_yield_wksht - Sn 3(PO 4 4 6 Na 2 CO 3 3 Sn(CO 3 2 4...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
PERCENT YIELD WORKSHEET 1) Write a balanced equation for the reaction of tin (IV) phosphate with sodium carbonate to make tin (IV) carbonate and sodium phosphate. 2) If 36 grams of tin (IV) phosphate is mixed with an excess of sodium carbonate, how many grams of tin (IV) carbonate will form? 3) If 29.8 grams of tin (IV) carbonate are actually formed when this reaction goes to completion, what is the percent yield? 4) If 7.3 grams of sodium carbonate are used in the reaction and the result a 74.0% yield, how many grams of sodium phosphate will be formed?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Everett Community College does not discriminate on the basis of race, religion, creed, color, national origin, age, sex, marital status, disability, or veteran status. PERCENT YIELD WORKSHEET - SOLUTIONS 1) Write a balanced equation for the reaction of tin (IV) phosphate with sodium carbonate to make tin (IV) carbonate and sodium phosphate.
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Sn 3 (PO 4 ) 4 + 6 Na 2 CO 3 3 Sn(CO 3 ) 2 + 4 Na 3 PO 4 2) If 36 grams of tin (IV) phosphate is mixed with an excess of sodium carbonate, how many grams of tin (IV) carbonate will form? 36 g Sn 3 (PO 4 ) 4 x 1 mole Sn 3 (PO 4 ) 4 x 3 mole Sn(CO 3 ) 2 x 238.73 g Sn(CO 3 ) 2 736 g Sn 3 (PO 4 ) 4 1 mole Sn 3 (PO 4 ) 4 1 mole Sn(CO 3 ) 2 = = 35 g Sn(CO 3 ) 2 3) If 29.8 grams of tin (IV) carbonate are actually formed when this reaction goes to completion, what is the percent yield? 29.8 g Sn(CO 3 ) 2 35 g Sn(CO 3 ) 2 x 100 = 85% 4) If 7.3 grams of sodium carbonate are used in the reaction and the result a 74.0% yield, how many grams of sodium phosphate will be formed? 7.3 g Na 2 CO 3 x 1 mole Na 2 CO 3 x 4 mole Na 3 PO 4 x 163.94 g Na 3 PO 4 105.99 g Na 2 CO 3 1 mole Na 2 CO 3 1 mole Na 3 PO 4 = = 7.5 g Na 3 PO 4 theoretical (7.5 g Na 3 PO 4 ) (0.74) = 5.6 g Na 3 PO 4 actual...
View Full Document

This note was uploaded on 10/04/2010 for the course CHEM 145 taught by Professor Prezhdo during the Fall '08 term at University of Washington.

Page1 / 2

Percent_yield_wksht - Sn 3(PO 4 4 6 Na 2 CO 3 3 Sn(CO 3 2 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online