A 2.00 g sample of ammonia (NH3) reacts with 4.00 g of oxygen (O2) according to the equation 4NH3+5O2→4NO+6H2O. How much excess reactant remains after the reaction has stopped ( knowing that NH3 is the excess reactant )?

0.017 moles of ammonia remains after the reaction is stopped.

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

The first step is to convert the mass to moles, of each reactant:

2 g .  1mol/ 17g = 0.117 moles of NH₃

4 g . 1mol /32g = 0.125 moles of O₂

Ammonia, states the question, is the excess reactant so we can confirm it,

5 moles of oxygen need 4 moles of ammonia to react (by stoichiometry)

Then, 0.125 moles of oxygen may react to (0.125 . 4) / 5 = 0.1 moles

As we have 0.117 moles of ammonia and we need 0.1 moles.

(0.117 - 0.1) = 0.017 moles remains after the reaction is completed.

If we convert the moles to mass we have:

0.017 mol . 17 g /1mol = 0.289 g

i dont know

explanation:

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