Question

How many liters of phosphine \(\left(\mathrm{PH}_{3}\right)\) gas at STP could be made from \(30 \mathrm{~g}\) of calcium by use of the following sequence of reactions: \(3 \mathrm{Ca}+2 \mathrm{P} \rightarrow \mathrm{Ca}_{3} \mathrm{P}_{2}\) \(\mathrm{Ca}_{3} \mathrm{P}_{2}+6 \mathrm{HCl} \rightarrow 2 \mathrm{PH} 3+3 \mathrm{CaCl}_{2}\) (Molecular weights: \(\mathrm{Ca}=40, \mathrm{PH}_{3}=34\).)

Short answer

11.2 liters of phosphine gas can be produced from 30 g of calcium at STP using the given sequence of reactions.

Step-by-step solution

Convert grams of calcium to moles

We are given 30 g of calcium and the molecular weight of calcium is 40. To convert grams to moles, we can use the formula: Moles = (mass of the substance) / (molar mass of the substance) So, for calcium, we have: Moles of calcium = (30 g) / (40 g/mol) = 0.75 mol

Use stoichiometry to find moles of phosphine produced

We can use stoichiometry to find out the moles of phosphine produced by the reactions: 3 Ca + 2 P → Ca3P2 Ca3P2 + 6 HCl → 2 PH3 + 3 CaCl2 From the equation, we can see that: - 3 moles of calcium produce 1 mole of calcium phosphide (Ca3P2) - 1 mole of calcium phosphide (Ca3P2) produces 2 moles of phosphine (PH3) So, we can set up a ratio to determine the moles of phosphine produced: (2 moles of PH3) / (3 moles of Ca) = x moles of PH3 / 0.75 moles of Ca To solve for x, multiply both sides by 0.75 moles of Ca: x moles of PH3 = (2 moles of PH3) * (0.75 moles of Ca) / (3 moles of Ca) x moles of PH3 = 0.5 moles of PH3

Calculate the volume of phosphine gas at STP

Now that we have the moles of phosphine gas, we need to find its volume at standard temperature and pressure (STP). At STP, 1 mole of any gas occupies 22.4 L. So, we can use the following formula to find the volume of phosphine gas: Volume = (moles of the gas) * (volume occupied by 1 mole at STP) For phosphine, we have: Volume of PH3 = (0.5 moles of PH3) * (22.4 L/mol) Volume of PH3 = 11.2 L So, 11.2 liters of phosphine gas can be produced from 30 g of calcium at STP using the given sequence of reactions.

Key concepts

Stoichiometry

Stoichiometry is an essential aspect of chemistry that involves quantifying the relationships between reactants and products in a chemical reaction. It helps us calculate the amount of products formed from a given quantity of reactants based on the balanced chemical equations. In our original exercise, stoichiometry is used to determine the number of moles of phosphine (\(\text{PH}_3\)) produced from a specific amount of calcium (Ca).
Here’s how it’s applied:

• Identify the balanced chemical equation, which shows the molar relationships between reactants and products.
• Using the given mass of a reactant, calculate its moles using the formula: \[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
• Apply the mole ratio from the balanced equation to find moles of the desired product.

These steps allow us to transition from a measured physical quantity (mass) to a useful chemical quantity (moles), which can then be used to predict the amount of product formed.

Molecular Weights

Molecular weights, also known as molar masses, refer to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). They are crucial in stoichiometry for converting between the mass of a substance and the number of moles. This value is derived from the sum of the atomic masses of all atoms in a molecule.
In our problem, the molecular weights given are:

• For calcium (Ca), the molecular weight is 40 g/mol. This means one mole of calcium weighs 40 grams.
• For phosphine (\(\text{PH}_3\)), it is 34 g/mol, implying one mole of phosphine has a mass of 34 grams.

Utilize these weights to perform conversions crucial for stoichiometric calculations. For instance, knowing 30 grams of calcium converts into 0.75 moles is essential for progressing through stoichiometric steps.

Standard Temperature and Pressure

Standard temperature and pressure (STP) is a reference point used in gas calculations, defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atm. At STP, one mole of an ideal gas occupies 22.4 liters of volume. This concept simplifies gas volume calculations by providing a constant condition.
In the given exercise:

• The conversion of moles of phosphine to its volume in liters was made straightforward using STP as a reference.
• 0.5 moles of phosphine, calculated based on stoichiometry, translate into 11.2 liters of gas when multiplied by 22.4 L/mol (the molar volume at STP).

STP assumptions are pivotal in this context, as they enable quick and reliable volume estimations without needing detailed pressure and temperature measurements for each reaction.

Calcium Reactions

Calcium, an alkaline earth metal, is highly reactive and forms various compounds through reactions. In our exercise, calcium reacts with phosphorus to form calcium phosphide (\(\text{Ca}_3\text{P}_2\)), which subsequently reacts with hydrochloric acid to yield phosphine gas.

• First, 3 moles of calcium react with 2 moles of phosphorus to form one mole of calcium phosphide. This reaction shows how metals combine with non-metals to form binary compounds.
• Following this, each mole of calcium phosphide reacts with hydrochloric acid, undergoing a decomposition reaction to produce 2 moles of phosphine (\(\text{PH}_3\)) and byproducts like calcium chloride.

Understanding these reactions and the role of calcium helps illustrate the chemical processes leading to the production of phosphine, an important aspect of the stoichiometric exercise.