Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation: CH4+NH3+O2→HCN+H2O. A student has 8 g of methane and 10 g of ammonia in excess oxygen.

a. What is the balanced equation for this reaction?
b. Which reagent is limiting? Explain why.
c. How many grams of hydrogen cyanide will be formed?

a.  

b. is the limiting.

c. 13.5 g of hydrogen cyanide will be formed

Explanation:

To calculate the moles :

The balanced reaction will be :

According to stoichiometry :

2 moles of require  = 2 moles of

Thus 0.5 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.

As 2 moles of give = 2 moles of

Thus 0.5 moles of give =  of

Mass of

Thus 13.5 g of will be produced from the given masses of both reactants.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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