Based on the Law of Conservation of Mass, what mass of products form when baking soda decomposes?
[tex]$NaHCO _3 \rightarrow Na _2 CO _3+ H _2 O + CO _2$[/tex]
25.00 g
? g
Give your answer to the correct number of significant figures.
Answer (1)
Balance the chemical equation: 2 N a H C O 3 → N a 2 C O 3 + H 2 O + C O 2 .
Calculate the moles of N a H C O 3 : 84.007 g / m o l 25.00 g = 0.297594 moles.
Determine the moles of each product: N a 2 C O 3 , H 2 O , and C O 2 are each 0.148797 moles.
Calculate the total mass of products: 15.7708 + 2.6806 + 6.5485 = 24.9999 g . The final answer is 25.00 g.
Explanation
Understanding the Problem The problem is based on the Law of Conservation of Mass, which states that the total mass of reactants in a chemical reaction is equal to the total mass of the products. We are given the mass of the reactant, baking soda ( N a H C O 3 ), and we need to find the total mass of the products ( N a 2 C O 3 , H 2 O , and C O 2 ) after the decomposition reaction. The given equation is unbalanced, which needs to be addressed first.
Balancing the Chemical Equation First, we need to balance the chemical equation: 2 N a H C O 3 → N a 2 C O 3 + H 2 O + C O 2 This balanced equation shows that 2 moles of N a H C O 3 decompose to form 1 mole of N a 2 C O 3 , 1 mole of H 2 O , and 1 mole of C O 2 .
Calculating Molar Mass of Reactant Next, we calculate the molar mass of N a H C O 3 :
Molar mass of N a H C O 3 = 22.99 ( N a ) + 1.01 ( H ) + 12.01 ( C ) + 3 × 16.00 ( O ) = 84.007 g / m o l
Calculating Moles of Reactant Now, we calculate the number of moles of N a H C O 3 using the given mass (25.00 g): Moles of $NaHCO_3 = \frac{25.00 g}{84.007 g/mol} = 0.297594
Calculating Molar Masses of Products We calculate the molar masses of the products: Molar mass of N a 2 C O 3 = 2 × 22.99 + 12.01 + 3 × 16.00 = 105.988 g / m o l Molar mass of H 2 O = 2 × 1.01 + 16.00 = 18.015 g / m o l Molar mass of C O 2 = 12.01 + 2 × 16.00 = 44.009 g / m o l
Calculating Moles of Products From the balanced equation, 2 moles of N a H C O 3 produce 1 mole of each product. Therefore: Moles of N a 2 C O 3 = 2 0.297594 = 0.148797 Moles of H 2 O = 2 0.297594 = 0.148797 Moles of C O 2 = 2 0.297594 = 0.148797
Calculating Mass of Each Product Now, we calculate the mass of each product: Mass of N a 2 C O 3 = 0.148797 × 105.988 = 15.7708 g Mass of H 2 O = 0.148797 × 18.015 = 2.6806 g Mass of C O 2 = 0.148797 × 44.009 = 6.5485 g
Calculating Total Mass of Products Finally, we calculate the total mass of the products: Total mass of products = 15.7708 + 2.6806 + 6.5485 = 24.9999 g Rounding to four significant figures, the total mass of the products is 25.00 g.
Examples
The Law of Conservation of Mass is a fundamental principle in chemistry and has numerous real-world applications. For example, in environmental science, it is used to track pollutants in ecosystems. If a factory releases a certain mass of pollutants into a river, scientists can use the Law of Conservation of Mass to predict how the pollutants will disperse and transform over time, ensuring that the total mass of the pollutant remains constant, even as it changes form through chemical reactions or physical processes. This helps in assessing the impact of pollution and developing effective remediation strategies. Similarly, in pharmaceutical manufacturing, the law ensures that the mass of reactants used to synthesize a drug equals the mass of the final product, guaranteeing the efficiency and yield of the production process.
