part e: balance the following equations by adding coefficients in the blanks.\n1. what are the steps for…

part e: balance the following equations by adding coefficients in the blanks.\n1. what are the steps for balancing a chemical equation?\na. write un - balanced equation\nb. count atoms of each element\nc. multiply coefficients\nd. verify final equation\n2. __mg + __o₂ → __mgo\n3. __cucl₂ + __h₂s → __cus + __hcl\npart f: determine the unknown amount of a substance in a chemical reaction.\n4. 40 g of calcium reacts with 71 g of chlorine to produce __ g of calcium chloride.\n5. __ g of potassium reacts with 16 g of oxygen to produce 94 g of potassium oxide.\n6. 14 g of lithium reacts with __ g sulfur to produce 46 g of lithium sulfide.

part e: balance the following equations by adding coefficients in the blanks.\n1. what are the steps for balancing a chemical equation?\na. write un - balanced equation\nb. count atoms of each element\nc. multiply coefficients\nd. verify final equation\n2. __mg + __o₂ → __mgo\n3. __cucl₂ + __h₂s → __cus + __hcl\npart f: determine the unknown amount of a substance in a chemical reaction.\n4. 40 g of calcium reacts with 71 g of chlorine to produce __ g of calcium chloride.\n5. __ g of potassium reacts with 16 g of oxygen to produce 94 g of potassium oxide.\n6. 14 g of lithium reacts with __ g sulfur to produce 46 g of lithium sulfide.

Answer

Explanation:

Step1: Balance the Mg - O₂ - MgO equation

For the reaction $\text{Mg}+\text{O}_2\rightarrow\text{MgO}$, there are 2 oxygen atoms on the left - hand side. To balance the oxygen, we put a 2 in front of MgO. Then we need 2 Mg on the left - hand side. So the balanced equation is $2\text{Mg}+\text{O}_2\rightarrow2\text{MgO}$.

Step2: Balance the CuCl₂ - H₂S - CuS - HCl equation

For the reaction $\text{CuCl}_2+\text{H}_2\text{S}\rightarrow\text{CuS}+\text{HCl}$, there are 2 chlorine atoms in $\text{CuCl}_2$ and 2 hydrogen atoms in $\text{H}_2\text{S}$. We put a 2 in front of HCl to balance the hydrogen and chlorine atoms. The balanced equation is $\text{CuCl}_2+\text{H}_2\text{S}\rightarrow\text{CuS} + 2\text{HCl}$.

Step3: Solve for the mass of calcium chloride

The reaction between calcium (Ca) and chlorine ($\text{Cl}_2$) to form calcium chloride ($\text{CaCl}_2$) is $\text{Ca}+\text{Cl}_2\rightarrow\text{CaCl}_2$. According to the law of conservation of mass, the mass of the reactants equals the mass of the products. So the mass of calcium chloride is $40 + 71=111$ g.

Step4: Solve for the mass of potassium

According to the law of conservation of mass for the reaction of potassium (K) and oxygen ($\text{O}_2$) to form potassium oxide ($\text{K}_2\text{O}$), let the mass of potassium be $x$. We know that $x + 16=94$, so $x=94 - 16 = 78$ g.

Step5: Solve for the mass of sulfur

Let the mass of sulfur be $y$. According to the law of conservation of mass for the reaction of lithium (Li) and sulfur (S) to form lithium sulfide ($\text{Li}_2\text{S}$), we have $14 + y=46$, so $y=46 - 14=32$ g.

Answer:

  1. $2\text{Mg}+\text{O}_2\rightarrow2\text{MgO}$
  2. $\text{CuCl}_2+\text{H}_2\text{S}\rightarrow\text{CuS}+2\text{HCl}$
  3. 111
  4. 78
  5. 32