Thread: Conservation of Mass

1. Conservation of Mass

Suppose 12 g of natural gas combines with 48 g of oxygen in a flame. The chemical change produces 33 g of carbon dioxide. How many grams of water form?

33-48+12=27g

I know this but is there an equation for this

2. Re: Conservation of Mass

Originally Posted by Louisana1
33-48+12=27g
33 - 48 + 12 ≠ 27.

3. Re: Conservation of Mass

Conversion of Energy UnitsThe complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced?

Is there an equation for this..The instructor says answer is 2.14 kJ

Ok i figured out this

512cal x 4.184J/cal = 2.14

4. Re: Conservation of Mass

Convert 2.75 x 10^4 kJ to calories
How is this figured out. Worded crazy to me

answer is this says my ughh instructor 6.57 x 10^6 cal

5. Re: Conservation of Mass

Originally Posted by Louisana1
Convert 2.75 x 10^4 kJ to calories
How is this figured out. Worded crazy to me

answer is this says my ughh instructor 6.57 x 10^6 cal
How many groups of 4.184J (i.e., calories) are contained in 2.75 x 104kJ = 2.75 x 107J?

6. Re: Conservation of Mass

Originally Posted by Louisana1
Suppose 12 g of natural gas combines with 48 g of oxygen in a flame. The chemical change produces 33 g of carbon dioxide. How many grams of water form?

33-48+12=27g

I know this but is there an equation for this
Yep:

CH4 + 2O2 --> CO2 + 2H20

The molecular weights of this reaction can be calculated using molecular weights of the constituent atoms: H=1, C=12, and O=16:

(12+4) + 2(16x2) --> 12+2(16) + 2(2x1 + 16),
16 + 64 --> 44 + 36,
80 --> 80

If you multiply each of thee by 3/4 you get the proportions youi asked about:

12 + 48 = 33 + 27