Answer: the empirical formula is C3H4O3
Explanation:Please see attachment for explanation
2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)
First off.. not a chem board.. but n e way.
This is a limiting reagent problem.
set it up as a DA problem.(Dimension Analysis)
Start with what you want.
you want Grams of acrylonitrile (C3H3N)
so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)
(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)
solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g
Same setup for the two other reactants.
so i did it and for
oxygen I got 11.04 grams
and for Ammonia i got 15.29 grams
So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.
Both the other reactants are in excess.
rate brainliest pls
The percent A by mass for substance AB =<u> 75%</u>
<h3>Further explanation</h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements
Empirical formula is the mole ratio of compounds forming elements.
From Substance AB₂ is 60.0% A by mass.
Let's say that AB₂ mass = 100 gram, then
mass A = 60 gram
mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram
In compound AB:
Total mass = mass A + mass B
Total mass = 60 + 20 grams = 80 grams
Then the percentage of compound A = (60: 80) = 75%
<h3>Learn more</h3>
Grams of KO₂ needed to form O₂
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Keywords : percent mass, substance
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Less friction to stop the wheel from turning
Answer:
6,216.684 kilograms of sodium carbonate must be added to neutralize
of sulfuric acid solution.
Explanation:
Mass of sulfuric acid solution = 

Percentage mass of sulfuric acid = 95.0%
Mass of sulfuric acid = 

Moles of sulfuric acid = 

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.
Then 58,647.96 moles of sulfuric acisd will be neutralized by :
of sodium carbonate
Mass of 58,647.96 moles of sodium carbonate :

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg
6,216.684 kilograms of sodium carbonate must be added to neutralize
of sulfuric acid solution.