Problem 83 - Chemistry Matter and Change

Question

Phosphorus combines with hydrogen to form phosphine. In this reaction, \(123.9 \mathrm{~g}\) of phosphorus combines with excess hydrogen to produce \(129.9 \mathrm{~g}\) of phosphine. After the reaction, \(310 \mathrm{~g}\) 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

6 g of hydrogen was used; the initial mass was 316 g.

Step 1: Calculate Mass of Hydrogen Used

In the reaction, the final product is phosphine, which weighs 129.9 g. Given that phosphorus initially weighs 123.9 g, we need to determine the mass of hydrogen that combined with the phosphorus. Since the total mass of phosphine is the sum of phosphorus and hydrogen used, calculate the hydrogen used as follows:\[\text{Mass of hydrogen used} = 129.9\, \text{g (phosphine)} - 123.9\, \text{g (phosphorus)} = 6\, \text{g}\]

Step 2: Determine Initial Mass of Hydrogen

We are given that 310 g of hydrogen remains unreacted after the reaction. Since 6 g of hydrogen reacted with phosphorus, the initial mass of hydrogen is the sum of used and unreacted hydrogen. Therefore, calculate it as follows:\[\text{Initial mass of hydrogen} = 6\, \text{g (used)} + 310\, \text{g (unreacted)} = 316\, \text{g}\]

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances into new products by breaking and forming bonds. In the context of our exercise, phosphorus reacts with hydrogen to form phosphine. This is a simple synthesis reaction, where two elements (phosphorus and hydrogen) combine to form a compound (phosphine). The most important aspect of any chemical reaction is the rearrangement of atoms; they are neither created nor destroyed, only rearranged. Another crucial point to note is that for any chemical reaction, each element's atoms must maintain a balanced count on both the reactant and product sides.

• Reactants are the starting substances (in this case, phosphorus and hydrogen).
• Products are the substances formed (here, phosphine).

Reactions are typically depicted using a chemical equation, which uses symbols to represent the reactants and products involved. While balancing these equations, it is essential to ensure the number of atoms for each element is equal on both sides.

Mass Conservation

Mass conservation is a foundational concept in chemistry that states the total mass of reactants in a chemical reaction must equal the total mass of the products. This means mass is neither gained nor lost; it is conserved. In the given exercise, calculating how much hydrogen participated in forming phosphine, and how much was left unreacted, showcases this principle. Initially, you have 123.9 g of phosphorus, and together with some quantity of hydrogen, it forms 129.9 g of phosphine.

• The mass of phosphine (product) is equal to the sum of the mass of phosphorus (reactant) and the mass of hydrogen (reactant) used in the reaction.

In this exercise, the reactants' total mass before the reaction equals the products' total mass after the reaction. Understanding this principle helps you solve practical chemistry problems and reinforces the idea that chemical equations must be balanced to reflect mass conservation accurately.

Reaction Calculations

Reaction calculations are a crucial skill, allowing chemists to quantify the substances involved in reactions. When working with stoichiometry, you'd need to know how to work out amounts of reactants needed or products formed. In our case, you are tasked with finding out two main things: how much hydrogen reacted to form phosphine, and how much you had initially.

• First, determine the mass of hydrogen used by subtracting the mass of phosphorus from the total mass of phosphine: \[129.9 \, \text{g (phosphine)} - 123.9 \, \text{g (phosphorus)} = 6.0 \, \text{g (hydrogen used)}\]
• Then, add the mass of the unreacted hydrogen to the hydrogen used to find the initial mass of hydrogen: \[6.0 \, \text{g (used)} + 310.0 \, \text{g (unreacted)} = 316.0 \, \text{g (initial)}\]

Through these calculations, stoichiometry demonstrates how balanced equations guide you in understanding quantitative relationships in reactions, ensuring that you respect the law of conservation of mass.