A chemist reacted 61.33 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. Na + Cl2 → NaCl

If the percentage yield of the reaction is 92%, what is the actual yield? Show your work including, the use of stoichiometric calculations and conversion factors.

143.43 grams

Explanation:

First, balance the chemical equation.

2 Na  +  Cl₂  ⇒  2 NaCl

Since chlorine is in excess, sodium is the limiting reagent.  This means that you need to find out the stoichiometric relationship between sodium and sodium chloride.  According to the equation, for every 2 moles of sodium, you produce 2 moles of sodium chloride.  This makes the ratio 1:1 (2/2 = 1/1).

Next, you need to convert grams of sodium to moles using the molar mass.

(61.33 g) ÷ (22.99 g/mol) = 2.667 mol

Convert moles of sodium to moles of sodium chloride using the ratio.

(2.667 mol Na) × (1 mol NaCl/1 mol Na) = 2.667 mol NaCl

Convert moles of sodium chloride to grams using the molar mass.

(2.667 mol) × (58.44 g/mol) = 155.90 g

You have now found the theoretical yield.  To find the actual yield, multiply the theoretical yield by the percent yield.

155.90 × 0.92 = 143.43

The chemist made 143.43 g of sodium chloride.

3 1/4 %

explanation:

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.