Question

Consider the reaction of sodium with carbon and ammonia. 2Na + 2C + 2NH3= 2NaCN + 3H2 Determine the limiting reactant in a mixture containing 96.9 g of Na, 58.2 g of C, and 107 g of NH3. Calculate the maximum mass (in grams) of sodium cyanide, NaCN, that can be produced in the reaction.

Answer 1

moles of Na = 96.9 / 23 = 4.213 mol

moles of C = 58.2 / 12 = 4.85 mol

moles of NH3 = 107 / 17 = 6.29 mol

2 Na   + 2 C    + 2 NH3 ---------------> 2NaCN + 3H2

2 mol        2 mol      2 mol                              2                

4.213        4.85        6.29                              

here the moles of Na is less . so

Na is limting reagent

2 mol Na   -----------------> 2 mol NaCN

4.213 mol Na    --------------> ??

moles of NaCN = 4.213 mol

mass of NaCN = 4.213 x 49 = 206 g

mass of NaCN = 206 g