Calculating masses of reactants&products in chemical reactions

In summary: CuO has a molar mass of 79.55 g/mol.For Cu2O:2 x 63.546 g/mol Cu = 127.092 g/mol Cu1 x 15.999 g/mol O = 15.999 g/mol OTotal = 143.091 g/molFor CuO:1 x 63.546 g/mol Cu = 63.546 g/mol Cu1 x 15.999 g/mol O = 15.999 g/mol OTotal = 79.545 g/molTo find the percentage of copper atoms, we divide the molar mass of copper by the total molar mass of
  • #1
debee
1
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i can't do this home work. i REALLY suck at physics, this is why I am here. can someone help me pleasse? :)

1#to combust completely 7g of iron, 2L of oxygen is needed. for each of the following state if any unreacred iron rests after combustion and if it does how much:
a) 2g of iron and 500mL of oxygen
b) 1g of iron and 300mL of oxygen
c) 1.75g of iron and 500mL oxygen


2# there exists two oxides of copper. Copper (I) oxide - Cu2O, and copper (II) ocide - CuO. calculate the % of copper atoms in each.
 
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  • #2


Hi there, I understand that you are struggling with your physics homework. I would be happy to help you with this problem!

1) To solve this problem, we first need to set up a balanced chemical equation for the combustion of iron. The equation is as follows:

4Fe + 3O2 -> 2Fe2O3

This means that for every 4 moles of iron (Fe) that reacts, we need 3 moles of oxygen (O2). Using the molar mass of iron (Fe = 55.845 g/mol) and oxygen (O2 = 32 g/mol), we can convert the given mass and volume of iron and oxygen into moles.

a) For 2g of iron and 500mL of oxygen:
2g Fe x (1 mol Fe/55.845 g) = 0.0358 mol Fe
500mL O2 x (1 L/1000 mL) x (1 mol O2/22.4 L) = 0.0223 mol O2

Since we have more moles of oxygen than iron, all the iron will be used up in the reaction and there will be no leftover iron.

b) For 1g of iron and 300mL of oxygen:
1g Fe x (1 mol Fe/55.845 g) = 0.0179 mol Fe
300mL O2 x (1 L/1000 mL) x (1 mol O2/22.4 L) = 0.0134 mol O2

Again, we have more moles of oxygen than iron, so all the iron will be used up in the reaction and there will be no leftover iron.

c) For 1.75g of iron and 500mL of oxygen:
1.75g Fe x (1 mol Fe/55.845 g) = 0.0313 mol Fe
500mL O2 x (1 L/1000 mL) x (1 mol O2/22.4 L) = 0.0223 mol O2

In this case, we have slightly more moles of iron than oxygen, so all the oxygen will be used up in the reaction and there will be 0.009 mol Fe left over.

2) To calculate the percentage of copper atoms in each oxide, we first need to find the molar mass of each oxide. Cu2O
 
  • #3


I am sorry to hear that you are struggling with your physics homework. Calculating masses of reactants and products in chemical reactions can be a challenging task, but with some practice and understanding of the concepts, you can definitely improve your skills.

Firstly, let's address the question about combusting iron and oxygen. In this reaction, iron (Fe) reacts with oxygen (O2) to form iron oxide (Fe2O3). The balanced chemical equation for this reaction is:

4Fe + 3O2 --> 2Fe2O3

From this equation, we can see that for every 4 moles of iron, we need 3 moles of oxygen to completely combust. Using the molar masses of iron (55.85 g/mol) and oxygen (32 g/mol), we can calculate the masses of reactants and products.

a) 2g of iron and 500mL of oxygen (Assuming oxygen is at STP)
Moles of iron = 2 g / 55.85 g/mol = 0.036 moles
Moles of oxygen = (500 mL / 22.4 L/mol) x 1 mol/1,000 mL = 0.022 moles
Since the ratio of iron to oxygen is 4:3, we can see that there is an excess of oxygen in this reaction. Therefore, all of the iron will combust and there will be no unreacted iron left.

b) 1g of iron and 300mL of oxygen (Assuming oxygen is at STP)
Moles of iron = 1 g / 55.85 g/mol = 0.018 moles
Moles of oxygen = (300 mL / 22.4 L/mol) x 1 mol/1,000 mL = 0.013 moles
Again, there is an excess of oxygen in this reaction, so all of the iron will combust and there will be no unreacted iron left.

c) 1.75g of iron and 500mL oxygen (Assuming oxygen is at STP)
Moles of iron = 1.75 g / 55.85 g/mol = 0.031 moles
Moles of oxygen = (500 mL / 22.4 L/mol) x 1 mol/1,000 mL = 0.022 moles
Here, the ratio of iron to oxygen is close to
 

What is the purpose of calculating masses of reactants and products in chemical reactions?

The purpose of calculating masses of reactants and products in chemical reactions is to determine the amount of each substance involved in the reaction. This information is important for understanding the stoichiometry of the reaction, predicting the amount of product that will be formed, and ensuring that the reaction is balanced.

How do you calculate the mass of a reactant or product in a chemical reaction?

To calculate the mass of a reactant or product, you need to know the chemical equation for the reaction, the molar mass of each substance, and the number of moles of the substance involved. You can then use the formula mass = moles x molar mass to calculate the mass of the reactant or product.

What is the importance of balancing chemical equations in calculating masses of reactants and products?

Balancing chemical equations is important because it ensures that the same number of atoms of each element are present on both the reactant and product sides of the equation. This is necessary for accurate calculations of masses, as well as for understanding the stoichiometry of the reaction.

How does the Law of Conservation of Mass apply to calculating masses of reactants and products?

The Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction. This means that the total mass of the reactants must equal the total mass of the products. When calculating masses of reactants and products, this law ensures that the results are accurate and balanced.

What are some common sources of error in calculating masses of reactants and products?

Some common sources of error in calculating masses of reactants and products include inaccurate measurements, rounding errors, and incomplete or incorrect chemical equations. It is important to use precise and accurate measurements and to double-check all calculations and equations to minimize errors.

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