21. consider the following chemical reaction: n2+ o2 2 no if 10.0 g of n2 reacts with excess oxygen then how many grams of no can be formed? a) 10.7 g b) 21.4 g c) 32.9 g d) 42.8 g page 4 of 8

The correct answer is Option B.

Explanation:

To calculate the number of moles, we use the equation:

  ....(1)

Given mass of nitrogen gas = 10 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

The given chemical reaction follows:

As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of nitrogen oxide.

So, 0.357 moles of nitrogen gas will produce = of nitrogen oxide.

Now, calculating mass of nitrogen oxide by putting values in equation 1, we get:

Moles of nitrogen oxide = 0.714 mol

Molar mass of nitrogen oxide = 30 g/mol

Putting values in equation 1, we get:

Hence, the correct answer is Option B.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g