Follow the powerpoint as an example of how to calculate the mass of the barium phosphate product ( in filter paper)with your data, we will use only the Barium Chloride as the known or starting point. The amount of the barium phosphate product should be compared to your experimental lab mass to get the percent error. The calculation should be neat and well explained, submit a picture by Wednesday midnight The formula of the barium chloride is BaCl 2 2 H 2 0, a hydrate, and its molar mass is 244. After you have calculated your yield of barium phosphate from the 1.2 grams of barium chloride, you should do the calculation using sodium phosphate ( 1.7.grams) as the known. Study the other slides as examples with other reactions.

REACTION STOICHIOMETRY(II) FACTOR LABEL LOGIC 2 Na 3 PO 4(s) + 3 BaCl 2 (s)  6 NaCl (aq) + 1 Ba 3 (PO 4 ) 2(s) KNOWN 1.2 g BaCl 2 DATA ( ) 244 gBaCl 2 1 mol BaCl 2 ( 1 mol Ba 3 (PO 4 ) 2 3 mol BaCl 2 ( ) 600 g Ba 3 (PO 4 ) 2 1 mol Ba 3 (PO 4 ) 2 1) CALCULATE THE YIELD OF BARIUM PHOSPHATE IN GRAMS IF 1.2 GRAMS OF BARIUM CHLORIDE IS CONSUMED IN THE ABOVE BALANCED REACTION. OBJECTIVE 0.98 Ba 3 (PO 4 ) 2 (1) CONVERT KNOWN TO MOLES FROM GRAMS, GRAMS CANCEL. MOLAR MASS OF BaCl 2 IS 244 G/MOL. GIVES MOLES OF KNOWN MOLES = MASS MOLAR MASS (2) MOLE RATIO: RATIO MOLES OF OBJECTIVE TO MOLES OF KNOWN., USE COEFFICIENTS OF BALANCED REACTION. THIS STEP GIVES MOLES OF OBJECTIVE. = (3) CONVERT MOLES OF OBJECTIVE TO GRAMS. THIS STEP GIVES GRAMS OF OBJECTIVE, WHICH THE QUESTION CALLED FOR )])]

REACTION STOICHIOMETRY(II) FACTOR LABEL LOGIC 2 H 2(g) + O 2 (g)  2 H 2 0 (g) KNOWN 4.O g H 2 O ( ) 18 gH 2 O 1 mol H 2 O ( ) 2 mol H 2 O 1 mol O 2 ( ) 36 g O 2 1 mol H 2 O 1) CALCULATE THE YIELD OF OXYGEN IN GRAMS IF 4.0 GRAMS OF WATER IS CONSUMED IN THE ABOVE BALANCED REACTION. OBJECTIVE 16 g H 2 O CONVERT KNOWN TO MOLES FROM GRAMS, GRAMS CANCEL. MOLAR MASS OF O 2 IS 18 G/MOL. GIVES MOLES OF KNOWN. MOLES = MASS MOLAR MASS MOLE RATIO: RATIO MOLES OF OBJECTIVE TO MOLES OF KNOWN., USE COEFFICIENTS OF BALANCED REACTION. THIS STEP GIVES MOLES OF OBJECTIVE. = CONVERT MOLES OF OBJECTIVE TO GRAMS. THIS STEP GIVES GRAMS OF OBJECTIVE, WHICH THE QUESTION CALLED FOR

REACTION STOICHIOMETRY FACTOR LABEL LOGIC 2 CH 3 OH + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O ( ) 2 MOLE CH 2 OH 2 mol CO 2 ( ) 2) CALCULATE THE TOTAL MASS OF PRODUCTS FORMED FROM 1 MOLE OF METHANOL. OBJECTIVE 44 g CO 2 = KNOWN 1 MOLE CH 3 OH 1 mol CO g CO 2 KNOWN 1 MOLE CH 3 OH ( ) 4 mol H 2 O 2 MOLE CH 2 OH ( ) 18. g H 2 O 1 mol H 2 O = OBJECTIVE 36 g LH 2 O THE SUM OF BOTH PRODUCTS IS 80. g