Question

For each of the following balanced chemical equations, calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant. a. \(\operatorname{AgNO}_{3}(a q)+\mathrm{LiOH}(a q) \rightarrow \mathrm{AgOH}(s)+\mathrm{LiNO}_{3}(a q)\) b. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{CaCl}_{2}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{CaSO}_{4}(s)\) c. \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

The products' mass produced by complete reaction of 0.125 mole of the first reactant in each equation are: a. 15.61 g of AgOH b. 48.63 g of CaSO4 c. 14.44 g of CaCl₂, 5.50 g of CO2, and 2.26 g of H2O d. 11.00 g of CO2 and 9.12 g of H2O

Step-by-step solution

Calculate the mass of 0.125 moles of AgNO3

Use the molar mass of AgNO3 (Ag: 107.87 g/mol, N: 14.01 g/mol, O: 16.00 g/mol) to find the mass of 0.125 moles: Mass of AgNO3 = moles × molar mass = 0.125 moles × (107.87 + 14.01 + 3 × 16.00) g/mol = 21.34 g

Determine the stoichiometry

From the balanced equation, the mole ratio between AgNO3 and AgOH is 1:1.

Calculate the mass of AgOH produced

Since the stoichiometry is 1:1, 0.125 moles of AgOH will be produced. Calculate its mass using the molar mass of AgOH (Ag: 107.87 g/mol, O: 16.00 g/mol, H: 1.01 g/mol): Mass of AgOH = moles × molar mass = 0.125 moles × (107.87 + 16.00 + 1.01) g/mol = 15.61 g Answer: 15.61 g of AgOH will be produced. b. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{CaCl}_{2}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{CaSO}_{4}(s)\)

Calculate the mass of 0.125 moles of Al2(SO4)3

Use the molar mass of Al2(SO4)3 (Al: 26.98 g/mol, S: 32.07 g/mol, O: 16.00 g/mol) to find the mass of 0.125 moles: Mass of Al2(SO4)3 = moles × molar mass = 0.125 moles × (2 × 26.98 + 3 × (32.07 + 4 × 16.00)) g/mol = 28.42 g

Determine the stoichiometry

From the balanced equation, the mole ratio between Al2(SO4)3 and CaSO4 is 1:3.

Calculate the mass of CaSO4 produced

Since the stoichiometry is 1:3, 0.375 moles of CaSO4 will be produced. Calculate its mass using the molar mass of CaSO4 (Ca: 40.08 g/mol, S: 32.07 g/mol, O: 16.00 g/mol): Mass of CaSO4 = moles × molar mass = 0.375 moles × (40.08 + 32.07 + 4 × 16.00) g/mol = 48.63 g Answer: 48.63 g of CaSO4 will be produced. c. \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\)

Calculate the mass of 0.125 moles of CaCO3

Use the molar mass of CaCO3 (Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol) to find the mass of 0.125 moles: Mass of CaCO3 = moles × molar mass = 0.125 moles × (40.08 + 12.01 + 3 × 16.00) g/mol = 12.55 g

Determine the stoichiometry

From the balanced equation, the mole ratio between CaCO3 and CaCl2, CO2, and H2O is 1:1:1:1.

Calculate the mass of CaCl2, CO2, and H2O produced

Since the stoichiometry is 1:1:1:1, 0.125 moles of CaCl2, CO2, and H2O will be produced. Calculate their masses using their molar masses (CaCl2: Ca: 40.08 g/mol, Cl: 35.45 g/mol; CO2: C: 12.01 g/mol, O: 16.00 g/mol; H2O: H: 1.01 g/mol, O: 16.00 g/mol): Mass of CaCl2 = 0.125 moles × (40.08 + 2 × 35.45) g/mol = 14.44 g Mass of CO2 = 0.125 moles × (12.01 + 2 × 16.00) g/mol = 5.50 g Mass of H2O = 0.125 moles × (2 × 1.01 + 16.00) g/mol = 2.26 g Answer: 14.44 g of CaCl₂, 5.50 g of CO2, and 2.26 g of H2O will be produced. d. \(2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\)

Calculate the mass of 0.0625 moles of C4H10 since first reactant value given is for 2 moles of C4H10

Use the molar mass of C4H10 (C: 12.01 g/mol, H: 1.01 g/mol) to find the mass of 0.0625 moles: Mass of C4H10 = moles × molar mass = 0.0625 moles × (4 × 12.01 + 10 × 1.01) g/mol = 4.94 g

Determine the stoichiometry

From the balanced equation, the mole ratio between two moles of C4H10 and CO2 and H2O formed is 2:8:10.

Calculate the mass of CO2 and H2O produced

For every mole of C4H10, there will be 4 moles of CO2 and 5 moles of H2O. Calculate their masses using their molar masses (CO2: C: 12.01 g/mol, O: 16.00 g/mol; H2O: H: 1.01 g/mol, O: 16.00 g/mol): Mass of CO2 = 0.25 moles × (12.01 + 2 × 16.00) g/mol = 11.00 g Mass of H2O = 0.3125 moles × (2 × 1.01 + 16.00) g/mol = 9.12 g Answer: 11.00 g of CO2 and 9.12 g of H2O will be produced.

Key concepts

Molar Mass Calculations

Molar mass calculations are fundamental to mastering stoichiometry, the branch of chemistry that deals with the relationships between reactants and products in a chemical reaction. Understanding molar mass is a stepping stone to converting between moles, the standard unit of amount in chemistry, and grams, which is more practical when measuring quantities of substances.

Imagine you have a Lego set with various blocks. Each type of block has a specific weight, and to know the weight of your creation, you need to know the weight of each type of block and how many of them you used. Similarly, each element on the periodic table has a unique atomic weight—its 'block' weight. To calculate the molar mass (essentially the weight of a mole) of a compound, you add up the atomic weights of each element in the compound according to its chemical formula. For instance, for AgNO3, we do this: molar mass of AgNO3 = mass of Ag + mass of N + 3 * mass of O, given the atomic weights Ag (107.87 g/mol), N (14.01 g/mol), and O (16.00 g/mol).

So, molar mass is essentially a conversion factor that helps us switch from counting molecules (in moles) to weighing them (in grams). It's an essential tool, much like knowing the exchange rate before converting dollars to euros.

Chemical Reaction Balancing

Balancing chemical reactions is akin to ensuring both sides of a scale are equally weighted, reflecting the law of conservation of mass—matter is neither created nor destroyed. It's a principle that dictates that atoms are simply rearranged in a reaction, not lost or gained.

In a balanced equation, the number of atoms for each element must be the same on both the reactants' and the products' side. To achieve this, we use coefficients—numbers placed in front of compounds—to adjust the number of atoms of each type until balance is achieved. Take, for example, the combustion of butane (C4H10): the equation starts as C4H10 + O2 -> CO2 + H2O. We need to determine the right coefficients to ensure we have the same number of carbon, hydrogen, and oxygen atoms on both sides. Getting this right allows us to understand the mole relationships between reactants and products, which is crucial for predicting the amounts of substances consumed and produced in the reaction.

Properly balanced equations are the backbone of stoichiometry; they function as a recipe that tells us how much of each ingredient we need and how much of the product we'll end up with.

Mole-to-Gram Conversions

Mole-to-gram conversions are the currency exchange of chemistry, allowing you to switch between the molecular world and the measurable, real world. When we say 'mole,' think of it as a specific number of particles, similar to a dozen eggs. One mole is Avogadro's number, 6.022 x 10^23, of particles—atoms, molecules, ions, or electrons. When we call for a mole of carbon, we mean 6.022 x 10^23 carbon atoms.

Here's where molar mass steps in again; it bridges moles and grams. Using the formula mass = moles * molar mass, you can determine the mass of a mole of any substance. For instance, if you're told to find the mass of 0.125 moles of CaCO3, you multiply 0.125 moles by the molar mass of CaCO3 (40.08 g/mol for Ca, 12.01 g/mol for C, and 16.00 g/mol for each of the 3 O atoms, totaling the molar mass).

In practical terms, mole-to-gram conversions are crucial. They tell a chemist how much of each substance to use in a reaction. Being adept at these conversions ensures precise experimentation and can help predict the expected yield of product from a given amount of reactant, which is vital for any chemical manufacturing process.