For this problem, identify the limiting reagent and calculate the grams of NaAl(

Question

For this problem, identify the limiting reagent and calculate the grams of NaAl(OH)4obtained in the reaction of 100 grams of Al2O3 with 75 grams of NaOH and 35 grams of water. If 130. grams of NaAl(OH)4 is actually produced, what is the % yield. The equations are not balanced. They were balanced in Exercise B. Use those coefficients to do these calculations.

Al2O3 + NaOH + H2O NaAl(OH)4

What is the limiting reagent? __________________

How many grams of NaAl(OH)4 will be produced? __________________

If 130 grams of NaAl(OH)4 are produced, what is the % yield?  

Explanation / Answer

The balanced reaction is given as :

Al2O3 + 2NaOH + 3H2O = 2NaAl(OH)4

Number of moles of Al2O3 = 100 /101.96 = 0.98 mol

Number of moles of NaOH = 75 / 39.997 = 1.875 mol

Number of moles of H2O = 35 / 18.01528 = 1.94 mol

0.98 mol of Al2O3 will require 2 x 0.98 = 1.96 mol of NaOH and 3 x 0.98 = 2.94 mol of H2O , so NaOH and H2O are the limiting reagents.

Now 3 moles of H2O makes 2 moles of NaAl(OH)4

So 1.94 mol will make : (1.94 x 2 ) / 3 = 1.295 mol of NaAl(OH)4

Grams of NaAl(OH)4 produced will be = 1.295 x 118 = 152.8 grams

% yield = ( 130 / 152.8) x 100

= 85 %

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