1. 32.5grams of NaOH (sodium hydroxide) reacts an excess of CO2 (carbon dioxide). 2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(1)
a.
What is the theoretical yield of Na2CO3 in grams? 43.1g Na2CO3
b.
If 16.45 g of Na2CO3 is formed, what is the percent yield? 38.2%

a. 43.1 g

b. 38.2%

Given

32.5 grams of NaOH

Required

The theoretical yield of Na₂CO₃

The percent yield

Solution

Reaction

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)

mol NaOH :

= mass : MW

= 32.5 : 40 g/mol

= 0.8125

mol Na₂CO₃ from the equation :

= 1/2 x mol NaOH

= 1/2 x 0.8125

= 0.40625

a.

Mass Na₂CO₃ :

= mol x MW Na₂CO₃

= 0.40625 x 106 g/mol

= 43.0625≈43.1 g

b. % yield = (actual/theoretical) x 100%

%yield = 16.45/43.1 x 1005

%yield = 38.17%≈38.2%

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

it is the most commonly accepted one