Fe2O3 + 3C → 2Fe + 3CO; how many grams of C are needed to react with 16.5 g Fe2O3? Chemistry Assignment Homework Help

User Generated

wreylsynhe

Science

Description

1. In the following reaction:  Fe2O3 + 3C → 2Fe + 3CO; how many grams of C are needed to react with 16.5 g Fe2O3?

2. Give an example of a physical change and chemical change.

 

3.  A sample of nitrogen (N2) has a volume of 50.0 L at a pressure of 760 mmHg. What is the volume of the gas when the pressure changes to 1500 mmHg if there is no change in temperature.

4. Calculate the molarity (M) of 4.0 g of KOH in 2.0 L of solution.

5.  Indicate whether the following solutions are acidic, basic or neutral.

  a. [H3O+] = 6.0 x 10-12 M

  b. [H3O+] = 1.4 x 10-4 M


User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Explanation & Answer

Dear student,Please find enclosed a doc file with a detailed explanation of the solutions to the 5 questions.

1. In the following reaction: Fe2O3 + 3C → 2Fe + 3CO, how many grams of C
are needed to react with 16.5 grams of Fe2O3?
In order to solve this question you need to focus on the concept of stoichiometry. Thus,
according to the coefficients in the balanced equation you can easily see that 1 mol of
Fe2O3 reacts with 3 moles of carbon. Since this proportion must be kept constant we
can use it to calculate the amount of carbon given to react with a specific amount of
Fe2O3 (16.5 grams in this case).
The last thing that you need to consider is the relationship between moles and grams.
Thus, the number of moles of any compound are calculated by dividing its mass by the
molar weight (the mass of 1 mole of compound). This molar weight is calculated as the
sum of the weights for each single atom present in the molecule. To calculate them,
you should check up the atomic weights of iron (Fe, 56 g/mol), carbon (C, 12 g/mol)
and oxygen (O, 15 g/mol) in a periodic table.
Taking this into account, the solution is calculated as:
1 𝑚𝑜𝑙 �...


Anonymous
Awesome! Made my life easier.

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4

Related Tags