Question

1. a student reacts 1.005 g al with 50 ml of 4m naoh and an excess of h₂so₄ in solution to produce 15.20 g of sodium alum.

a. calculate the molar mass of sodium alum, naal(so₄)₂·12h₂o.
b. what is the theoretical yield of sodium alum? (the molar ratio between al and sodium alum is 1 to 1.)
c. what is the percent yield of sodium alum?

Explanation:

Step1: Calculate molar - mass of NaAl(SO₄)₂·12H₂O

Molar mass of Na: $22.99\ g/mol$, Al: $26.98\ g/mol$, S: $32.07\ g/mol$, O: $16.00\ g/mol$, H: $1.01\ g/mol$.
$M = 22.99+26.98 + 2\times32.07+8\times16.00+12\times(2\times1.01 + 16.00)$
$M=22.99 + 26.98+64.14 + 128.00+12\times(2.02 + 16.00)$
$M=22.99 + 26.98+64.14 + 128.00+12\times18.02$
$M=22.99 + 26.98+64.14 + 128.00+216.24$
$M = 453.35\ g/mol$

Step2: Calculate moles of Al

$n_{Al}=\frac{m_{Al}}{M_{Al}}$, where $m_{Al}=1.005\ g$ and $M_{Al}=26.98\ g/mol$.
$n_{Al}=\frac{1.005}{26.98}\ mol\approx0.03725\ mol$
Since the molar - ratio of Al to sodium alum is 1:1, the theoretical moles of sodium alum $n_{theo}=n_{Al}=0.03725\ mol$.

Step3: Calculate theoretical yield of sodium alum

$m_{theo}=n_{theo}\times M$, where $M = 453.35\ g/mol$ and $n_{theo}=0.03725\ mol$.
$m_{theo}=0.03725\times453.35\ g\approx16.89\ g$

Step4: Calculate percent yield of sodium alum

Percent yield $=\frac{m_{actual}}{m_{theo}}\times100\%$, where $m_{actual}=15.20\ g$ and $m_{theo}=16.89\ g$.
Percent yield $=\frac{15.20}{16.89}\times100\%\approx90.00\%$

Answer:

a. $453.35\ g/mol$
b. $16.89\ g$
c. $90.00\%$