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Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction and allows for the calculation of the amount of a different element/compound in the same reaction. Obviously the place to start is….
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Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction and allows for the calculation of the amount of a different element/compound in the same reaction. Obviously the place to start is…
Predict and write a balanced chemical equation. Start a RR track and convert the given information to moles (if not already as moles).
Predict and write a balanced chemical equation. • Start a RR track and convert the given information to moles (if not already as moles). • Use the molar mass (from the PT) to change grams → moles
Predict and write a balanced chemical equation. • Start a RR track and convert the given information to moles (if not already as moles). • Use the molar mass (from the PT) to change grams → moles • Use 6.022 x 1023 to change atoms/molecules/formula units → moles
Predict and write a balanced chemical equation. • Start a RR track and convert the given information to moles (if not already as moles). • Use the molar mass (from the PT) to change grams → moles • Use 6.022 x 1023 to change atoms/molecules/formula units → moles • Use 22.4 L (for gases only) to change L → moles
Predict and write a balanced chemical equation. • Start a RR track and convert the given information to moles (if not already as moles). • Use the molar mass (from the PT) to change grams → moles • Use 6.022 x 1023 to change atoms/molecules/formula units → moles • Use 22.4 L (for gases only) to change L → moles • For water and dilute solutions, 1 mL = 1 g
Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation • Mole # of unknown • Mole # of known
Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation • Mole # of unknown • Mole # of known • 4. Convert the moles of the unknown to the desired unit (see possible conversion factors in step 2)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1:
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2 Find the moles of phosphoric acid Find the mass of phosphoric acid
Find the formula units of phosphoric acid Find the moles of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the formula units of phosphoric acid Find the moles of potassium phosphate Find the mass of potassium phosphate
Find the mass of the hydrogen Find the liters of hydrogen
Find the mass of the hydrogen Find the liters of hydrogen
Find the mass of the hydrogen Find the liters of hydrogen
Find the mass of the hydrogen Find the liters of hydrogen