7.16 pts.) gas. The balanced reacts with sodium metal to form sodium amide (NaNH

Question

7.16 pts.) gas. The balanced reacts with sodium metal to form sodium amide (NaNH2 and hydrogen 2 NH. chemical equation for this reaction is given below: Molar mass g mo 2 NaNH. Na 22.99 NH3 17.03 A student starts with 22.8 g of ammonia gas and 352 g of sodium metal. NaNH2 39.02. 2.016 (a) Determine the limiting reagent. (b) Find the mass in grams of NaNH2 produced (c) The mass of NaNH2 you determined in part a will be your theoretical yield. What is the actual yield of NaNH. in grams if the percent yield of this experiment was 92.8%? (d) What mass in grams of which reactant is left over?

Explanation / Answer

Answer –

In this problem there are given, mass of reactants.

Mass of NH3 = 22.8 g , mass of Na = 35.2 g

Reaction –

a)Limiting reactant –

First we need to calculate the moles of reactants

Moles of NH3 = 22.8 g / 17.03 g.mol-1

                       = 1.34 moles

Moles of Na = 35.2 g / 22.99 g.mol-1

                     = 1.53 moles

Now we need to calculate moles of NaNH2 from each reactant –

First from the NH3

2 moles of NH3 = 2 moles of NaNH2

So, 1.34 moles of NH3 = ?

= 1.34 moles of NaNH2

From Na metal –

2 moles of Na = 2 moles of NaNH2

So, 1.53 moles of Na = ?

= 1.53 moles of NaNH2

So moles of NaNH2 is lowest from the moles of NH3, so limiting reactant is NH3.

b) Mass of NaNH2 produced –

From the limiting reactant the moles of NaNH2 = 1.34 moles

so theoretical yield of NaNH2 = 1.34 moles * 39.02 g/mol

                                                = 52.24 g

c) In this one we are given, percent yield = 92.8 %, we already calculated theoretical yield

we need to calculate actual yield, we know formula

percent yield = actual yield / theoretical yield *100 %

actual yield = percent yield * theoretical yield / 100 %

                    = 92.8 % * 52.24 g / 100 %

                    = 48.48 g

d) Mass of excess reactant –

We know excess reactant is Na , so first we need to calculate how much moles of Na used

From the given balanced equation –

2 moles of NH3 = 2 moles of Na

So, 1.34 moles of NH3 = ?

= 1.34 moles of Na

So 1.34 moles of Na used and remaining are 1.53-1.34 = 0.192 moles

So mass of excess reactant , Na = 0.192 moles * 22.99 g/mol

                                                  = 4.42 g

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