Problem 83 - Chemistry Matter and Change

Question

Phosphorus combines with hydrogen to form phosphine. In this reaction, 123.9 g of phosphorus combines with excess hydrogen to produce 129.9 g of phosphine. After the reaction, 310 g of hydrogen remains unreacted. What mass of hydrogen is used in the reaction? What was the initial mass of hydrogen before the reaction?

Step-by-Step Solution

Answer

The mass of hydrogen used in the reaction is 6 g, and the initial mass of hydrogen before the reaction was 316 g.

Step 1: Identify the Mass of Phosphorus and Phosphine

We're given that 123.9 g of phosphorus reacts with excess hydrogen to produce 129.9 g of phosphine.

Step 2: Calculate the Mass of Hydrogen Used

To find the mass of hydrogen used in the reaction, subtract the mass of phosphorus from the mass of phosphine. This will give us the mass of hydrogen that has combined with phosphorus. Mass of Hydrogen used = Mass of Phosphine - Mass of Phosphorus Mass of Hydrogen used = 129.9 g - 123.9 g Mass of Hydrogen used = 6 g The mass of hydrogen used in the reaction is 6 g.

Step 3: Find the Mass of Unreacted Hydrogen

We are given that 310 g of hydrogen remains unreacted after the reaction.

Step 4: Calculate the Initial Mass of Hydrogen

Now, add the mass of hydrogen used in the reaction to the mass of unreacted hydrogen to find the initial mass of hydrogen before the reaction. Initial Mass of Hydrogen = Mass of Hydrogen used + Mass of Unreacted Hydrogen Initial Mass of Hydrogen = 6 g + 310 g Initial Mass of Hydrogen = 316 g The initial mass of hydrogen before the reaction was 316 g.

Key Concepts

Chemical Reaction Calculation

Understanding chemical reaction calculations involves balancing chemical equations and determining the quantities of reactants and products.

When phosphorus combines with hydrogen to produce phosphine, we start by expressing this reaction in a balanced chemical equation. Although the specific equation isn't provided in the original exercise, a balanced equation is typically required for more complex reactions. Balancing the equation ensures that the same number of each type of atom is present on both sides, aligning with the conservation of mass.

Once the reaction is expressed and balanced, calculations can proceed to determine the quantities involved. In the given problem, the masses of phosphorus and phosphine were used to calculate the mass of hydrogen that reacted. Even without the chemical equation, the principle that the mass of reactants equals the mass of products stands, leading to the calculation outlined in the steps.

Mass Conservation Law

The law of conservation of mass states that matter cannot be created or destroyed in an isolated system. This principle is crucial in stoichiometry because it allows us to predict the outcomes of chemical reactions.

In the context of the exercise, the law of mass conservation is applied to find the mass of hydrogen that reacted with phosphorus. The mass of phosphine produced is the sum of the masses of phosphorus and hydrogen that reacted. In more detailed problems, knowledge of the conservation of mass can assist in balancing chemical equations by ensuring that the mass of each element is the same on both sides of the equation. Moreover, the law helps us to deduce other valuable information, such as the initial mass of hydrogen, by adding the mass of hydrogen that reacted with the unreacted remainder.

Molar Mass Conversion

Molar mass conversion is an essential concept in stoichiometry and provides a bridge between the mass of a substance and the number of particles or moles. It is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

Understanding molar mass is critical for converting the weights of reactants and products to moles, which can then be used to derive stoichiometric relationships in a balanced chemical equation. While molar mass was not directly used in the problem described, more complex stoichiometry problems may require converting given masses to moles using the molar mass of the substances involved. By doing so, students can work with the mole ratios from the balanced equation, enabling them to predict the amounts of products formed or reactants consumed in a chemical reaction.