Question

Silver bromide, \(\mathrm{AgBr}\), used in photography, may be prepared from \(\mathrm{AgNO}_{3}\) and \(\mathrm{NaBr}\). Calculate the weight of each required for producing \(93.3 \mathrm{lb}\) of \(\mathrm{AgBr} .(1 \mathrm{lb}=454 \mathrm{~g}\) )

Short answer

To produce 93.3 lb of silver bromide (AgBr), 38,268.4 g of silver nitrate (AgNO3) and 23,307.5 g of sodium bromide (NaBr) are required.

Step-by-step solution

Convert weight of AgBr into grams

Since 1 lb = 454 g, let's convert the given weight of AgBr (93.3 lb) into grams: \( 93.3~lb \times \frac{454~g}{1~lb} \) Now calculate the weight of AgBr in grams: Weight of AgBr = 93.3 * 454 = 42,334.2 g

Write the balanced chemical equation for the reaction

The reaction between silver nitrate (AgNO3) and sodium bromide (NaBr) to form silver bromide (AgBr) and sodium nitrate (NaNO3) is given by: AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq)

Calculate the molar masses of reactants and product

Calculate the molar masses of AgNO3, NaBr, and AgBr: Molar mass of AgNO3: Ag - 107.87 g/mol N - 14.01 g/mol O3 - 3 * 16.00 g/mol = 48.00 g/mol Total = 107.87 + 14.01 + 48.00 = 169.88 g/mol Molar mass of NaBr: Na - 22.99 g/mol Br - 79.90 g/mol Total = 22.99 + 79.90 = 102.89 g/mol Molar mass of AgBr: Ag - 107.87 g/mol Br - 79.90 g/mol Total = 107.87 + 79.90 = 187.77 g/mol

Determine the mass of each reactant required

To find the mass of each reactant required, we will use stoichiometry: For AgNO3: Since 1 mole of AgNO3 produces 1 mole of AgBr, from the balanced chemical equation, moles of AgNO3 = moles of AgBr = \(\frac{42,334.2~g~AgBr}{187.77~g/mol}\) Now, convert moles of AgNO3 to grams: Mass of AgNO3 = moles of AgNO3 × molar mass of AgNO3 For NaBr: Since 1 mole of NaBr produces 1 mole of AgBr, from the balanced chemical equation, moles of NaBr = moles of AgBr = \(\frac{42,334.2~g~AgBr}{187.77~g/mol}\) Now, convert moles of NaBr to grams: Mass of NaBr = moles of NaBr × molar mass of NaBr Now calculate the mass of AgNO3 and NaBr required: Mass of AgNO3 = (moles of AgBr) * (169.88 g/mol) = (42,334.2 / 187.77) * 169.88 = 38,268.4 g Mass of NaBr = (moles of AgBr) * (102.89 g/mol) = (42,334.2 / 187.77) * 102.89 = 23,307.5 g The weight of AgNO3 required is 38,268.4 g and the weight of NaBr required is 23,307.5 g for producing 93.3 lb of AgBr.

Key concepts

Molar Mass Calculation

Molar mass is a key concept in stoichiometry and represents the mass of one mole of a substance. It is calculated by summing up the atomic masses of all atoms in a chemical formula. For example, to find the molar mass of silver bromide (AgBr), we combine the atomic masses of silver (Ag) and bromine (Br).
Ag has an atomic mass of approximately 107.87 g/mol, while Br is around 79.90 g/mol. Therefore, the molar mass of AgBr is the sum: 107.87 + 79.90 = 187.77 g/mol.
This value helps us in determining how much of each substance is involved in a chemical reaction. Calculating molar masses is a foundational step before performing any stoichiometric calculations.

Balanced Chemical Equation

A balanced chemical equation ensures that the number of each type of atom is the same on both sides of the reaction. Consider our example involving silver nitrate (AgNO3) and sodium bromide (NaBr), which react to form silver bromide (AgBr) and sodium nitrate (NaNO3).
The equation can be written as: AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq), indicating that one mole of AgNO3 reacts with one mole of NaBr to produce one mole of AgBr and one mole of NaNO3.
Balancing chemical equations is vital because it reflects the Law of Conservation of Mass. This law states that matter cannot be created or destroyed in a chemical reaction, ensuring consistency in mass and atom counts across the equation.

Silver Bromide (AgBr) Production

Silver bromide (AgBr) is a compound commonly used in photographic film. It can be synthesized through a straightforward chemical reaction between silver nitrate (AgNO3) and sodium bromide (NaBr).
The balanced equation, AgNO3 + NaBr → AgBr + NaNO3, highlights that the reaction proceeds with all reactants and products in a one-to-one molar ratio. This stoichiometric relationship allows us to predict how much of each reactant is needed to produce a desired amount of AgBr.
Understanding the production process of AgBr is essential for applications in photography, where precise amounts of silver bromide are required for optimal imaging results.

Reactant Mass Calculation

Calculating the mass of reactants needed for a specific product in a chemical reaction involves understanding the mole concept and molar masses. Let's examine how to calculate the amounts needed to produce a specific weight of silver bromide (AgBr).
Initially, convert the target mass of AgBr from pounds to grams. Here, 93.3 lb is 42,334.2 g. Using the molar mass of AgBr (187.77 g/mol), determine the moles of AgBr: 42,334.2 g / 187.77 g/mol.
Then, based on the balanced equation, the moles of reactants (AgNO3 and NaBr) will equal the moles of AgBr. Finally, convert these moles back into grams using their respective molar masses (AgNO3 = 169.88 g/mol and NaBr = 102.89 g/mol). This conversion gives the masses: AgNO3 is 38,268.4 g and NaBr is 23,307.5 g.
This calculation ensures that the correct quantities of reactants are used to achieve the desired product amount.