b) mass (in grams) of carbon in 58.5 g of sodium acetate

Explanation:

Step1: Determine molar mass of sodium acetate ($C_2H_3NaO_2$)

The molar masses of $C = 12.01\ g/mol$, $H=1.01\ g/mol$, $Na = 22.99\ g/mol$, $O = 16.00\ g/mol$. $M(C_2H_3NaO_2)=2\times12.01 + 3\times1.01+22.99 + 2\times16.00=82.03\ g/mol$

Step2: Determine the number of moles of sodium acetate

$n=\frac{m}{M}$, where $m = 58.5\ g$ and $M = 82.03\ g/mol$. $n=\frac{58.5\ g}{82.03\ g/mol}\approx0.713\ mol$

Step3: Determine moles of carbon in sodium acetate

In one - mole of $C_2H_3NaO_2$, there are 2 moles of carbon. So, moles of carbon in $0.713\ mol$ of $C_2H_3NaO_2$ is $n_{C}=2\times0.713\ mol = 1.426\ mol$

Step4: Calculate mass of carbon

$m_{C}=n_{C}\times M_{C}$, where $M_{C}=12.01\ g/mol$ and $n_{C}=1.426\ mol$ $m_{C}=1.426\ mol\times12.01\ g/mol\approx17.1\ g$

Answer:

$17.1\ g$