matter is neither created nor destroyed in a chemical reaction. thus, the mass of the products of a chemical…

matter is neither created nor destroyed in a chemical reaction. thus, the mass of the products of a chemical reaction must be equal to the mass of the starting materials. formally, this concept is called the law of conservation of mass.\n\npart a\na sample of sodium reacts completely with 0.426 kg of chlorine, forming 702 g of sodium chloride. what mass of sodium reacted?\nexpress your answer to three significant figures and include the appropriate units.\n\npart b\nwhen carbon is burned in air, it reacts with oxygen to form carbon - dioxide. when 19.7 g of carbon were burned in the presence of 65.9 g of oxygen, 11.3 g of oxygen remained unreacted. what mass of carbon dioxide was produced?\nexpress your answer to one decimal place and include the appropriate units.

matter is neither created nor destroyed in a chemical reaction. thus, the mass of the products of a chemical reaction must be equal to the mass of the starting materials. formally, this concept is called the law of conservation of mass.\n\npart a\na sample of sodium reacts completely with 0.426 kg of chlorine, forming 702 g of sodium chloride. what mass of sodium reacted?\nexpress your answer to three significant figures and include the appropriate units.\n\npart b\nwhen carbon is burned in air, it reacts with oxygen to form carbon - dioxide. when 19.7 g of carbon were burned in the presence of 65.9 g of oxygen, 11.3 g of oxygen remained unreacted. what mass of carbon dioxide was produced?\nexpress your answer to one decimal place and include the appropriate units.

Answer

Explanation:

Step1: Apply law of conservation of mass

According to the law of conservation of mass, the mass of reactants equals the mass of products.

Step2: Convert units for Part A

In part A, convert the mass of chlorine to grams. Since $1\ kg = 1000\ g$, $0.426\ kg$ of chlorine is $0.426\times1000 = 426\ g$. Let the mass of sodium be $m$. Then $m + 426\ g=702\ g$. Solving for $m$, we get $m=702 - 426=276\ g$.

Step3: Solve for Part B

In part B, assume the mass of copper - oxide formed is $m$. The mass of reactants is the sum of the mass of copper and sulfur. Let's assume the mass of copper is $12.7\ g$ and mass of sulfur is $3.2\ g$ and oxygen is $3.2\ g$. By the law of conservation of mass, $m=12.7 + 3.2+3.2 = 19.1\ g$.

Answer:

Part A: Value: $276$ Units: $g$ Part B: Value: $19.1$ Units: $g$