For the reaction, calculate how many moles of the product form when 0.032 mol of k completely reacts. assume that there is more than enough of the other reactant. 4k(s)+o2(g)→2k2o(s) express your answer using two significant figures.

0.016 moles of product will obtain.

Explanation:

Given data:

Moles of K = 0.032 mol

Moles of product = ?

Solution:

Chemical equation:

4K + O₂ →  2K₂O

Now we will compare the moles of K with K₂O.

          K          :           K₂O

           4         :            2

        0.032    :           2/4 ×0.032 = 0.016 mol

0.016 moles of product will obtain.

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