Unit 4: Topic 5 - Stoichiometry Flashcards
How do we calculate reactant and product amounts?
Dimensional Analysis: a mathematical approach to convert from one unit to another using conversion factors
To convert from grams to moles of a substance:
For example, how many moles do you have from 50 grams of sodium (Na)?
50g Na x (1 mol Na / 23g Na) = 2.2 mol Na
The conversion factor (1 mol Na / 23g Na) shows that there are 23g of Na for every mole of Na (molar mass). Multiplying this by the given total mass of Na (50g) leaves us with the amount of moles of Na.
Relatedly, to convert from moles to grams of a substance, multiply the number of moles by the molar mass (g/mol) to find the amount of substance in grams (mass).
To convert from a reactant amount to a product amount, set up multiple conversion factors (make sure other units cancel out to leave you with the desired unit). A typical setup looks like this: grams of reactant x (1 mol reactant/molar mass of reactant in g) x (mol product/mol reactant) x (molar mass of product in g/1 mol product). Notice how the units cancel out to leave you with grams of product.
Avogadro’s Number is the number of units in one mole of any substance, equal to 6.02214076 x 10²³.
Convert the following to moles:
a) 35.00g of CF₂Cl₂
b) 100.0mg of iron(II) sulfate
To find the molar mass of a compound:
1) Use the chemical formula to determine the number of each type of atom present in the compound.
2) Multiply the atomic weight of each element by the number of atoms of that element present in the compound.
3) Add everything together and put units of grams/mole after the number.
a) 0.2893 mol CF₂Cl₂
molar mass of CF₂Cl₂: 12.0g/mol C + 2(19.0g/mol F) + 2(35.5g/mol Cl) = 121.0g/mol CF₂Cl₂
35.00g CF₂Cl₂ x (1 mol CF₂Cl₂ / 121.0g CF₂Cl₂) = 0.2893 mol CF₂Cl₂
b) 6.579 x 10⁻⁴ mol FeSO₄
molar mass of iron(II) sulfate (FeSO₄): 55.9g/mol Fe + 32.1g/mol S + 4(16.0g/mol O) = 152.0g/mol FeSO₄
100.0mg = 0.1000g
0.1000g FeSO₄ x (1 mol FeSO₄ / 152.0g FeSO₄) = 0.0006579 mol FeSO₄ = 6.579 x 10⁻⁴ mol FeSO₄
Phosphine gas reacts with oxygen according to the following equation:
4PH₃(g) + 8O₂(g) -> P₄O₁₀(s) + 6H₂O(g)
Calculate:
a) the mass of tetraphosphorous decaoxide produced from 12.43 mol of phosphine.
b) the mass of PH₃ required to form 0.739 mol of steam.
c) the mass of oxygen gas required to react with 20.50g of phosphine.
molar mass of PH₃: 34g/mol PH₃
molar mass of P₄O₁₀: 284g/mol P₄O₁₀
molar mass of O₂: 32g/mol O₂
a) 882.5g P₄O₁₀
12.43 mol PH₃ x (1 mol P₄O₁₀ / 4 mol PH₃) x (284g P₄O₁₀ / 1 mol P₄O₁₀) = 882.5g P₄O₁₀
b) 16.75g PH₃
0.739 mol H₂O x (4 mol PH₃ / 6 mol H₂O) x (34g PH₃ / 1 mol PH₃) = 16.75g PH₃
c) 38.59g O₂
20.50g PH₃ x (1 mol PH₃ / 34g PH₃) x (8 mol O₂ / 4 mol PH₃) x (32g O₂ / 1 mol O₂) = 38.59g O₂
How do the coefficients of a chemical equation help us determine the mole ratio?
Mole Ratio: relates the amounts of moles of any two substances in a chemical reaction
The coefficients in front of each substance in a balanced chemical equation indicate the number of moles of each reactant that are reacted together to produce the number of moles of each product. Atoms of substances must be conserved during the chemical reaction.
For example, consider the chemical formation of water: 2H2(g) + O2(g) -> 2H2O(l)
The coefficients tell you that 2 moles of hydrogen gas (H2) reacted with 1 mole of oxygen gas (O2) produces 2 moles of water (H2O). We can write the relationship between hydrogen and water as a mole ratio: 2 mol H2 / 2 mol H2O. Similarly, we can write the relationship between oxygen and water as a mole ratio: 1 mol O2 / 2 mol H2O.
How can we use the limiting reactant (reagent) of a reaction to determine the mass of product formed?
Limiting Reactant: The reactant that is consumed first in a chemical reaction and therefore limits how much product can be formed.
1) Determine moles of each reactant.
2) Divide the moles of each reactant by its coefficient from the balanced equation.
3) Determine the limiting reactant. The one with the smallest number from step 2 is the limiting reactant.
4) Use the moles of limiting reactant to determine the moles and mass of the product.
Consider the reaction of 80.0g of copper with 50.0g of oxygen.
4Cu + O₂ -> 2Cu₂O
a) What is the limiting reagent?
b) What mass of product is formed?
a) Cu is the limiting reagent.
80.0g Cu x (1 mol Cu / 63.6g Cu) = 1.26 mol Cu
1.26 mol Cu / 4 mol Cu = 0.315
50.0g O₂ x (1 mol O₂ / 32g O₂) = 1.56 mol O₂
1.56 mol O₂ / 1 mol O₂ = 1.56
b) 90.2g Cu₂O
1.26 mol Cu x (2 mol Cu₂O / 4 mol Cu) x (143.1g Cu₂O / 1 mol Cu₂O) = 90.2g Cu₂O
Calculate the molarity of the solution if a 5.623g sample of NaHCO₃ is dissolved in enough water to make 250.0 mL of solution.
0.2678 M
molar mass of NaHCO₃: 84g/mol NaHCO₃
250.0 mL = 0.2500 L
5.623g NaHCO₃ x (1 mol NaHCO₃ / 84g NaHCO₃) = 0.0669 mol NaHCO₃
0.0669 mol NaHCO₃ / 0.2500 L = 0.2678 M
Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: 100.0 mL of 0.30 M AlCl₃, 50.0 mL of 0.60 M MgCl₂, or 200.0 mL of 0.40 M NaCl?
100.0 mL of 0.30 M AlCl₃
AlCl₃: 0.100 L x (0.30 M / L) = 0.030 mol AlCl₃
0.030 mol AlCl₃ x 3 Cl = 0.090 mol Cl⁻
MgCl₂: 0.0500 L x (0.60 M / L) = 0.030 mol MgCl₂
0.030 mol MgCl₂ x 2 Cl = 0.060 mol Cl⁻
NaCl: 0.2000 L x (0.40 M / L) = 0.080 mol NaCl
0.080 mol NaCl x 1 Cl = 0.080 mol Cl⁻
