Lecture : Quantitative Information from Balanced Equations. 2 H 2 + (1) O 2 → 2 H 2 O 2 moles 1 mole 2 moles

3.6: Quantitative Information from Balanced Equations. The coefficients in a balanced equation represent both the relative numbers of molecules involved in the reaction, AND THE RELATIVE NUMBERS OF MOLES, and therefore the relative masses: 2 H 2 (g)+O 2 (g) →2 H 2 O (l) 2 molecules + 1molecule2 molecules 2 moles 1 mole 2 moles = 2 x (2.0) g g→2 x (18.0) g

Coefficients in a balanced equation: The coefficients in a balanced equation as written mean ‘molecules’. So we have 2 H 2 (g)+O 2 (g) →2 H 2 O (l) 2 molecules + 1molecule2 molecules We can multiply through the whole equation by Avogadro’s number, and then we have moles: 2 x x x x x x = 2 moles 1 mole 2 moles

Weights of products and reactants in a balanced equation: We can thus work out how many grams of water will be produced by burning given amounts of H 2 and O 2 together. Example: The combustion of butane: 2 C 4 H 10 (l) + 13 O 2 (g) → 8 CO 2 (g) + 10 H 2 O(l) 2 moles 13 moles 8 moles 10 moles 2 x 58.0 g 8 x 44.0 g M. Wt. C 4 H 10 M. Wt. CO 2

How many grams of CO 2 will be obtained by burning 1.00 g of C 4 H 10 ? 1) Grams reactant → moles Conversion factor: 1 = 1 mol/58.0 g 1.00 g x 1mol/58 g = mol 2) Moles reactant → Moles product 2 moles C 4 H 10 → 8 moles CO 2 moles CO 2 = x 8/2 = moles 3) Moles product → grams product moles CO 2 = mol x 44.0 g/1 mol grams of CO 2 = 3.03 g coefficients from balanced equation

3.7. Limiting Reactants. Cheese sandwiches: (Ch = slice of cheese, Bd = slice of bread) 2 Bd + 1 Ch=Bd 2 Ch If we have 12 Ch and 8 Bd, how many sandwiches can we make? Obviously, only 4, with 8 Ch left over. We are limited in this case by the amount of Bd we have. Bd in this case is the limiting ingredient.

Limiting Reactants. An analogous situation occurs with chemical reactions. Consider the reaction: 2 H 2 (g)+O 2 (g)→2 H 2 O(l) 2 mol+1 mol2 mol If we have exactly 2 mol of H 2 and 1 mol of O 2, then we can make 2 mol of water. But what if we have 4 mol of H 2 and 1 mol of O 2. Now we can make only 2 mol H 2 O with 2 mol H 2 left over. In this case the O 2 is the limiting reagent. The limiting reagent is the one with nothing left over.

Multiplying an equation through by a common multiple: We can multiply all the coefficients in a balanced equation by any multiple, and it still has the correct ratios of moles. Thus, if we have: Zn(s) + 2HCl(aq)  ZnCl 2(aq) +H 2 (g) 1 mole 2 moles 1 mole + 1 mole If we have 2 moles of Zn(s), this gives: (x 2) 2 moles 4 moles 2 moles 2 moles or if we have 0.5 moles Zn(s) we have: (x 0.5) 0.5 moles 1 mole 0.5 moles 0.5 moles

How do we solve limiting reagent problems? We try both reagents out as the possible limiting reagent. Consider the reaction of H 2 and N 2 to give NH 3, and assume we have 3.0 mol N 2 and 6.0 mol H 2. Which is the limiting reagent? We have the balanced equation: N 2 (g)+3 H 2 (g)→2 NH 3 (g) 1 mol3 mol2 mol Factor = moles N 2 we have moles N 2 in equation = 3.0 mol N mol N 2 = 3.0 (multiply all coefficients in balanced equation by this factor)

Multiply all coefficients by factor (x 3): N 2 (g)+3 H 2 (g)→2 NH 3 (g) 1 mol 3 mol 2 mol 3 mol3 x 3 = 9 mol 3 x 2 = 6 mol Try N 2 as limiting reagent: 3 mol N 2 requires how many moles H 2 ? =3 x 3 = 9 mol We only have 6 mol H 2, so H 2 is the limiting reagent.

Example: Consider the following reaction: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol 3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO3(aq) 1 mol 6 mol How much Ba 3 (PO 4 ) 2 can be formed if we have in the solutions 3.50 g sodium phosphate and 6.40 g barium nitrate?

Step 1. Convert to moles: First work out numbers of Moles: Na 3 PO 4 = 3.50 g x 1 mol = mol 164 g Ba(NO 3 ) 2 = 6.40 g x 1 mol= mol 261 g

Step 2. Guess limiting reagent (it doesn’t matter if you guess wrong): Guess that Ba(NO 3 ) 2 is the limiting reagent: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol Factor = mol (moles we have) 3 mol(moles in equation) =

Guessing which reagent is the limiting reagent: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol We notice that we have only slightly more moles of Ba(NO 3 ) 2 ( moles) than Na 3 PO 4 ( moles), but the Ba(NO 3 ) 2 has a coefficient of 3 in the balanced equation, while that of Na 3 PO 4 is only 2. In general go for the substance with the higher coefficient and higher molecular mass.

Step 3. Multiply all the moles in the equation by the factor: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol times factor ( ): 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol or mol mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol mol mol

Step 4. Compare required moles of Na 3 PO 4 with moles we previously calculated ( moles) 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq) = mol mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) mol mol We have moles of Na 3 PO 4, so we guessed right, Ba(NO 3 ) 2 is limiting reagent (if we guessed wrong calculate new factor with other reagent and repeat calculations). bigger than

How much Ba 3 (PO 4 ) 2 can be formed? We now have an equation with the correct number of moles of each reactant if the limiting reactant is moles of Ba(NO 3 ) 2 : 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  mol mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) mol mol So we need to calculate how many grams of Ba 3 (PO 4 ) 2 there are in moles.

Step 5. Convert moles of Ba 3 (PO 4 ) 2 to grams: We now need g of Ba 3 (PO 4 ) 2. F. Wt. = 3 x x x 16.0 = g/mol Need grams, so we have mol x601.9 g= 4.92 g. 1 mol

Practice Exercise: Zn metal (2.00 g) plus solution of AgNO 3 (2.50 g) reacts according to: Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) Ag(s) 1 mol2 mol Which is the limiting reagent? How much Zn will be left over?

Step 1. Convert to moles: Zn = g/mol AgNO 3 = (3 x 16) = g/mol Zn = 2.0 g x1 mol= mol g AgNO 3 = 2.50 g x 1 mol = mol g

Step 2. Guess limiting reagent Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) Ag(s) 1 mol2 mol In this case it seems clear that AgNO 3 must be the limiting reagent, because the equation says we must have 2 mols of AgNO 3 for each mol of Zn(s), but in fact we have more moles of Zn(s).

We can check this by using the AgNO 3 to calculate a factor that can be used to multiply through the equation. Factor = /2 = Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) Ag(s) 1 mol2 mol We in fact have mol of Zn, which is more than the mol required, so AgNO 3 is clearly the limiting reactant.

How much Zn will be left over? We actually have mol of Zn, but require only mol. We will therefore have mol – mol = mol = 65.4 x mol = 1.52 g left over (At. Wt. Zn)

Percent Yield: Theoretical yields: The quantity of product that forms if all of the limiting reagent reacts is called the theoretical yield. Usually, we obtain less than this, which is known as the actual yield. Percent yield =actual yieldx 100 Theoretical yield

Problem: 10.4 g of Ba(OH) 2 was reacted with an excess of Na 2 SO 4 to give a precipitate of BaSO 4. If the reaction actually yielded 11.2 g of BaSO 4, what is a) the theoretical yield of BaSO 4 and b) what is the percentage yield of BaSO 4 ? The balanced equation for the reaction is: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq)

Step 1. Convert to moles: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq) 1 mole1 mole 1 mole 2 moles Moles Ba(OH) 2 : Mol. Mass Ba(OH) 2 = x ( ) = g/mol Moles = 10.4 g x 1 mol = moles g

Step 2. Work out how much BaSO 4 will be formed: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq) 1 mole1 mole 1 mole 2 moles moles When it says that one reagent is in excess, that means we do not have to worry about that reagent, and the other one is the limiting reagent, in this case the BaSO 4. We see that 1 mole of Ba(OH) 2 will produce 1 mole of BaSO 4. Our factor is thus , and we will get moles of BaSO 4.

Calculation of theoretical yield: Convert moles of BaSO 4 to grams: Formula mass BaSO 4 = = 4 x 16.0 =235.3 g/mol No of grams of BaSO 4 expected is: moles x g =14.29 g 1 mole (theoretical yield)

Convert actual yield to percentage yield: Percent yield =actual yield x 100 % Theoretical yield =11.2 g x100 % g =78.4 % yield