How Do You Calculate the Amount of Each Ingredient Needed to Make Gunpowder?

  • Thread starter Thread starter b14him88
  • Start date Start date
  • Tags Tags
    Gun Powder
AI Thread Summary
To calculate the grams of each ingredient needed to make gunpowder, start by using stoichiometry based on the balanced equation 2KNO3 + 3C + S → 3CO2 + K2S + N2. Determine the molar mass of each reactant: potassium nitrate (KNO3), carbon (C), and sulfur (S). For a total of 10 grams, use the ratios from the balanced equation to find the proportionate amounts of each substance. Remember that the total mass of reactants must equal the total mass of products, and not all ingredients will convert to the desired products. Apply these principles to calculate the exact grams needed for your chemistry class project.
b14him88
Messages
2
Reaction score
0
I need some help on how to find out how many grams of each substance i need to make gun powder for my Chemisty class. I have the equation its 2KNO3+3C+S -->3CO2 + K2S+ N2 I know that there are 10 grams but i have to figure out how many grams are for each substance. I'd just like someone point me in the right direction. I'm not sure what to do first. I know i need to use stoikiometry to figure it out. but if someone could just give me a little advice on what to do first. i would appreciate it a lot. thank you. as far as i know i have balanced the equation correctly but if someone sees a mistake, let me know please. thanks again.

-Janelle
 
Physics news on Phys.org
b14him88 said:
I need some help on how to find out how many grams of each substance i need to make gun powder for my Chemisty class. I have the equation its 2KNO3+3C+S -->3CO2 + K2S+ N2 I know that there are 10 grams but i have to figure out how many grams are for each substance. I'd just like someone point me in the right direction. I'm not sure what to do first. I know i need to use stoikiometry to figure it out. but if someone could just give me a little advice on what to do first. i would appreciate it a lot. thank you. as far as i know i have balanced the equation correctly but if someone sees a mistake, let me know please. thanks again.

-Janelle

I do hope this is purely an academic exercise and you are not trying to actually make this stuff. A balanced equation tells you how many atoms of each element are involved in the reaction. If you have exactly the right ratios for the number of atoms, you will have a complete reaction. Since atoms of different elements have different weights, knowing the ratios of the numbers is not the same as knowing the ratios of the weights, but if you know the number ratios, and the weights of each atom, you can figure out the ratios of weights.

For example, the reaction for water is

2H2+O2 -->2H2O

4 hydrogen atoms and 2 oxygen atoms combine to form two water molecules. In atomic mass units (amu), the weight of hydrogen is about 1 and the weight of oxygen is about 16. If A is some unspecified multiplying factor measured in grams per amu, you would need to have

A*4*1amu + A*2*16amu = grams of water formed.

You could sove this equation for A. Then you could find the weight of each element needed for the reaction.
 
Oh geez.. you totally confused me with everything u said. and yes i am making gun powder @ school in my chemistry class- that is if i figure out the equation. the amu stuff..yeah idk what ur talking about. all i know if that we have to figure out the moles, the g/mol, and the mass. maybe u could put it in lamens terms so i can understand? thanks for the attempt though- sorry that I am dumb lol.
 
b14him88 said:
Oh geez.. you totally confused me with everything u said. and yes i am making gun powder @ school in my chemistry class- that is if i figure out the equation. the amu stuff..yeah idk what ur talking about. all i know if that we have to figure out the moles, the g/mol, and the mass. maybe u could put it in lamens terms so i can understand? thanks for the attempt though- sorry that I am dumb lol.

In terms of moles and grams per mole then- The reaction for water is

2H2+O2 -->2H2O

2 hydrogen molecules and 1 oxygen molecule combine to form two water molecules. In terms of moles, 2 moles of hydrogen molecules and one mole of oxygen molecules combine to form two moles of water molecules. If A is some unspecified multiplying factor, you would need to have

A*(2 moles hydrogen)*(2 grams/mole) + A*(1 mole oxygen)*(32 grams/mole) = A*(2 moles water)*(18 grams/mole) = # grams of water formed.

If A = 1 you would get 4 grams hydrogen + 32 grams oxygen --> 36 grams of water (2 moles)

If A = 1/2 you would get 18 grams of water (1 mole)
etc.

If you wanted 10 grams of water, you would find the value of A that will give you 10 grams of water. That would be

A = 10 grams/(4 grams + 32 grams) = 5/18

So you would need (5/18)*2 moles of hydrogen or 5/9 moles and (5/18)*1 moles of oxygen or 5/18 moles and the result would be

(5/18)*(2 moles hydrogen)*(2 grams/mole) + (5/18)*(1 mole oxygen)*(32 grams/mole) = 10 grams of water

So now you know how many moles of hydrogen you need, and you can figure out how many grams that is because you know there are 2 grams per mole. You also know how many moles of oxygen you need and you can figure out haw many grams that is.

Follow the same steps for your reaction, and you will arrive at the number of grams of each compound needed. In your problem you have to also take into consideration that the product of the reaction includes other compounds besides the one you are trying to produce, so not all of the ingredients are turned into desired products. You should be able to figure out how to extend the water example to the more general case.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top