Advertisements
Advertisements
Question
In a reaction \[\ce{x + y + z2 -> xyz2}\] identify the Limiting reagent if any, in the following reaction mixture.
200 atoms of x + 200 atoms of y + 50 molecules of z2
Solution
Reaction: \[\ce{x + y + z2 -> xyz2}\]
200 atoms of x + 200 atoms of y + 50 molecules of z2.
According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give the product.
In case 200 atoms of x, 200 atoms of y react with 50 molecules of z2 (4 part) i.e. 50 molecules of z2 react with 50 atoms of x and 50 atoms of y.
Hence z is the limiting reagent.
APPEARS IN
RELATED QUESTIONS
When 22.4 litres of H2 (g) is mixed with 11.2 litres of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
In a reaction \[\ce{x + y + z2 -> xyz2}\] identify the Limiting reagent if any, in the following reaction mixture.
1 mol of x + 1 mol of y + 3 mol of z2
In a reaction \[\ce{x + y + z2 -> xyz2}\] identify the Limiting reagent if any, in the following reaction mixture.
50 atoms of x + 25 atoms of y + 50 molecules of z2
In a reaction \[\ce{x + y + z2 -> xyz2}\] identify the Limiting reagent if any, in the following reaction mixture.
2.5 mol of x + 5 mol of y + 5 mol of z2