# Chapter 12 Stoichiometry Goal 4 Calculate quantities of reactants and products needed in chemical rxns using balanced chemical equations.

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Chapter 12

Stoichiometry Goal 4 Calculate quantities of reactants and products needed in chemical rxns using balanced chemical equations.

Using an Equation as a Recipe 4I4Ingredients: Frame, Seat, Wheels, Handlebar, Pedals 4B4Balanced equation: 1F + 1S + 2W + 1H + 2P 1 FSW 2 HP 2

How many pedals are needed to make 128 bicycles? = 256 pedals 128 FSW 2 HP 2 1 FSW 2 HP 2 2P 1F + 1S + 2W + 1H + 2P 1 FSW 2 HP 2

Examining a Balanced Chemical Equation 2H 2 S(g) + 3O 2 (g) 2SO 2 (g) + 2H 2 O(g) 4242 moles of H 2 S reacts with 3 moles of O 2 to produce 2 moles of SO 2 and 2 moles of H 2 O. 4 moles of H 2 S reacts with ___moles of O 2 4 ____ moles of O 2 are needed to produce 7 moles of H 2 O 4H4How to solve these is on the next slide. 6 10.5

= 6 mol O 2 4 mol H 2 S 2 mol H 2 S 3 mole O 2 Mole-Mole Ratio Problems = 10.5 mol O 2 7mol H 2 O 2 mol H 2 O 3 mole O 2 The mol-mol ratio comes from the coefficients in the balanced equation. In the balance equation, O 2 has a coefficient of 3, H 2 O has a coefficient of 2, and H 2 S has a coefficient of 2.

Mass-Mass Calculation ·W·W·W·Write the balanced chemical equation ¸L¸L¸L¸Label above the equation what you are given and what you are trying to find. ¹C¹C¹C¹Convert grams of your given to moles using molar mass ºMºMºMºMultiply by the mole-mole ratio from the balanced chemical equation »C»C»C»Convert moles of unknown to grams

Bringing It All Together 4 How many grams of ammonia ( ) should be produced when 6.00g nitrogen react with excess hydrogen? 4 Solve: Write the balanced equation with the information labeled above it. 6.00 g excess ? g N 2 + 3H 2 2 NH 3 NH3

Bringing It All Together 4 Convert given to moles: 4 Multiply by the mole ratio: = 0.428 mol NH 3 = 0.214 mol N 2 6. 00g N 2 28 g N 2 1 mole N 2 0.214 mol N 2 1 mol N 2 2 mol NH 3

Wrapping It Up Convert moles of unknown to grams: = 7.28g NH 3 Summary: When 6.00g of nitrogen(N 2 ) reacts with an excess of Hydrogen (H 2 ), 7.28g of ammonia (NH 3 ) should be produced. 0.428 mol NH 3 1 mol NH 3 17g NH 3

Now You Try! What mass of aluminum is required to completely react with 7.80g oxygen in a synthesis reaction?

Write the balanced equation 4Al + 3O 2 2Al 2 O 3

Label your given and unknown information 4Al + 3O 2 2Al 2 O 3 7.80 g Since you were not given info about this compound, don’t worry about it and don’t use it.

NEXT?... 4 Convert given to moles: 4 Multiply by the mole ratio: 7.80g O 2 32g O 2 1 mol O 2 = 0.244mol O 2 0.244mol O 2 3 mol O 2 4 mol Al = 0.325 mol Al

AND FINALLY!! Convert moles of unknown to grams: 0.325 mol Al 1 mol Al 27g Al = 8.78g Al 8.78g of aluminum (Al) should be required to react completely with7.80g of oxygen (O 2 ) in the synthesis of aluminum oxide (Al 2 O 3 ).

Mass-Volume Calculation 4 Write the balanced chemical equation 4 Label above the equation what you are given and what you are trying to find. 4 Convert grams of your given to moles using molar mass 4 Multiply by the mole-mole ratio from the balanced chemical equation 4 Convert moles of unknown to liters (a.k.a dm 3 ) using molar volume (22.4L/ 1mol)

Bringing It All Together 4 How many liters of oxygen are necessary for the combustion of 134g of magnesium, assuming the reactions occurs at STP? 4 Solve: Write the balanced equation with the information labeled above it. 134 g ? L 2Mg + O 2 2 MgO

Bringing It All Together 4 Convert given to moles: 4 Multiply by the mole ratio: = 5.514 mol Mg = 2.757 mol O 2 134g Mg 24.3 g Mg 1 mole Mg 5.514 mol Mg 2 mol Mg 1 mol O 2

Wrapping It Up Convert moles of unknown to liters: = 61.8 L O 2 61.8 L of O 2 are required to completely combust 134g Mg in the synthesis of MgO. 2.757 molO 2 1 mol O 2 22.4 LO 2

Volume-Mass Calculation 4 Write the balanced chemical equation 4 Label above the equation what you are given and what you are trying to find. 4 Convert liters of your given to moles using molar volume 4 Multiply by the mole-mole ratio from the balanced chemical equation 4 Convert moles of unknown to grams using molar mass

Bringing It All Together 4 Find the mass of sulfur (S 8 ) required to react with oxygen to produce 2.47L of sulfur dioxide gas at STP. 4 Solve: Write the balanced equation with the information labeled above it. ? g 2.47L S 8 + 8 O 2 8 SO 2

Bringing It All Together 4 Convert given to moles: 4 Multiply by the mole ratio: = 0.1103 mol SO 2 = 0.01378 mol S 8 2.47L SO 2 22.4 L SO 2 1 mole SO 2 0.1103 mol SO 2 8 mol SO 2 1 mol S 8

Wrapping It Up Convert moles of unknown to grams: = 3.53 g S 8 3.53g of sulfur (S 8 ) are required to produce 2.47 L of sulfur dioxide (SO 2 ). 0.01378 mol S 8 1 mol S 8 256g S 8

Volume-Volume Calculations 4 Write the balanced chemical equation 4 Label above the equation what you are given and what you are trying to find. 4 Convert liters of your given to moles using molar volume (22.4L/1mol) 4 Multiply by the mole-mole ratio from the balanced chemical equation 4 Convert moles of unknown to liters using molar volume (22.4L/1 mol)

Bringing It All Together 4 What volume of H 2 S gas is needed to produce 14.2 L of water at STP? 4 Solve: Write the balanced equation with the information labeled above it. ?L 14.2L 2H 2 S + 3O 2 2 SO 2 + 2 H 2 O

Bringing It All Together 4 Convert given to moles: 4 Multiply by the mole ratio: = 0.6339 mol H 2 O = 0.6339 mol H 2 S 14.2 L H 2 O 22.4 L H 2 O 1 mole H 2 O 0.6339 mol H 2 O 2mol H 2 O 2 mol H 2 S

Wrapping It Up Convert moles of unknown to liters: = 14.2 L H 2 S 14.2 L of H 2 S are required to produce 14.2 L H 2 O. 0.6993 mol H 2 S 1 mol H 2 S 22.4 LH 2 S

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