13. calculate the molar mass of the following compounds.\na. n₂o₃\nb. mg₃n₂\nc. agno₃\n14. how many\na. fe…

13. Calculate the molar mass of the following compounds

a. $N_2O_3$

Explanation:

Step1: Find molar mass of N and O

The molar mass of N is $M_N = 14.01\ g/mol$, and of O is $M_O=16.00\ g/mol$.

Step2: Calculate molar mass of $N_2O_3$

$M_{N_2O_3}=2\times M_N + 3\times M_O=2\times14.01\ g/mol+3\times16.00\ g/mol = 28.02\ g/mol + 48.00\ g/mol=76.02\ g/mol$

b. $Mg_3N_2$

Explanation:

Step1: Find molar mass of Mg and N

The molar mass of Mg is $M_{Mg}=24.31\ g/mol$, and of N is $M_N = 14.01\ g/mol$.

Step2: Calculate molar mass of $Mg_3N_2$

$M_{Mg_3N_2}=3\times M_{Mg}+2\times M_N=3\times24.31\ g/mol + 2\times14.01\ g/mol=72.93\ g/mol+28.02\ g/mol = 100.95\ g/mol$

c. $AgNO_3$

Explanation:

Step1: Find molar mass of Ag, N and O

The molar mass of Ag is $M_{Ag}=107.87\ g/mol$, of N is $M_N = 14.01\ g/mol$ and of O is $M_O = 16.00\ g/mol$.

Step2: Calculate molar mass of $AgNO_3$

$M_{AgNO_3}=M_{Ag}+M_N+3\times M_O=107.87\ g/mol+14.01\ g/mol+3\times16.00\ g/mol=107.87\ g/mol + 14.01\ g/mol+48.00\ g/mol=169.88\ g/mol$

14.

a. How many Fe atoms are in 0.256 mol Fe?

Explanation:

Step1: Use Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}\ atoms/mol$.

Step2: Calculate number of Fe atoms

$N = n\times N_A$, where $n = 0.256\ mol$. So $N=0.256\ mol\times6.022\times 10^{23}\ atoms/mol=1.54\times 10^{23}\ atoms$

b. How many Kr atoms are in 3.87 g Kr?

Explanation:

Step1: Find molar mass of Kr

The molar mass of Kr is $M_{Kr}=83.80\ g/mol$.

Step2: Calculate number of moles of Kr

$n=\frac{m}{M}$, where $m = 3.87\ g$ and $M = 83.80\ g/mol$. So $n=\frac{3.87\ g}{83.80\ g/mol}=0.0462\ mol$.

Step3: Calculate number of Kr atoms

$N=n\times N_A$, where $N_A = 6.022\times 10^{23}\ atoms/mol$. So $N = 0.0462\ mol\times6.022\times 10^{23}\ atoms/mol=2.78\times 10^{22}\ atoms$

c. How many $AlCl_3$ formula - units are in 6.17 mol $AlCl_3$?

Explanation:

Step1: Use Avogadro's number

Avogadro's number $N_A=6.022\times 10^{23}\ formula - units/mol$.

Step2: Calculate number of $AlCl_3$ formula - units

$N=n\times N_A$, where $n = 6.17\ mol$. So $N=6.17\ mol\times6.022\times 10^{23}\ formula - units/mol = 3.71\times 10^{24}\ formula - units$

d. How many $NaIO_3$ formula - units are in 8.58 g $NaIO_3$?

Explanation:

Step1: Find molar mass of $NaIO_3$

The molar mass of Na is $M_{Na}=22.99\ g/mol$, of I is $M_{I}=126.90\ g/mol$ and of O is $M_O = 16.00\ g/mol$. So $M_{NaIO_3}=M_{Na}+M_{I}+3\times M_O=22.99\ g/mol+126.90\ g/mol+3\times16.00\ g/mol=22.99\ g/mol+126.90\ g/mol + 48.00\ g/mol=197.89\ g/mol$.

Step2: Calculate number of moles of $NaIO_3$

$n=\frac{m}{M}$, where $m = 8.58\ g$ and $M = 197.89\ g/mol$. So $n=\frac{8.58\ g}{197.89\ g/mol}=0.0434\ mol$.

Step3: Calculate number of $NaIO_3$ formula - units

$N=n\times N_A$, where $N_A = 6.022\times 10^{23}\ formula - units/mol$. So $N=0.0434\ mol\times6.022\times 10^{23}\ formula - units/mol=2.61\times 10^{22}\ formula - units$

15.

a. What is the mass in grams of 6.58 mol Fe?

Explanation:

Step1: Find molar mass of Fe

The molar mass of Fe is $M_{Fe}=55.85\ g/mol$.

Step2: Calculate mass of Fe

$m=n\times M$, where $n = 6.58\ mol$ and $M = 55.85\ g/mol$. So $m=6.58\ mol\times55.85\ g/mol = 367\ g$

b. What is the mass in grams of $8.58\times 10^{28}$ atoms Kr?

Explanation:

Step1: Calculate number of moles of Kr

$n=\frac{N}{N_A}$, where $N = 8.58\times 10^{28}$ atoms and $N_A=6.022\times 10^{23}\ atoms/mol$. So $n=\frac{8.58\times 10^{28}\ atoms}{6.022\times 10^{23}\ atoms/mol}=1.425\times 10^{5}\ mol$.

Step2: Find molar mass of Kr

The molar mass of Kr is $M_{Kr}=83.80\ g/mol$.

Step3: Calculate mass of Kr

$m=n\times M$, where $n = 1.425\times 10^{5}\ mol$ and $M = 83.80\ g/mol$. So $m=1.425\times 10^{5}\ mol\times83.80\ g/mol=1.19\times 10^{7}\ g$

c. What is the mass in grams of 1.05 mol $AlCl_3$?

Explanation:

Step1: Find molar mass of $AlCl_3$

The molar mass of Al is $M_{Al}=26.98\ g/mol$ and of Cl is $M_{Cl}=35.45\ g/mol$. So $M_{AlCl_3}=M_{Al}+3\times M_{Cl}=26.98\ g/mol+3\times35.45\ g/mol=26.98\ g/mol + 106.35\ g/mol=133.33\ g/mol$.

Step2: Calculate mass of $AlCl_3$

$m=n\times M$, where $n = 1.05\ mol$ and $M = 133.33\ g/mol$. So $m=1.05\ mol\times133.33\ g/mol = 140\ g$

d. What is the mass in grams of $3.17\times 10^{18}$ formula units $NaIO_3$?

Explanation:

Step1: Calculate number of moles of $NaIO_3$

$n=\frac{N}{N_A}$, where $N = 3.17\times 10^{18}$ formula - units and $N_A=6.022\times 10^{23}\ formula - units/mol$. So $n=\frac{3.17\times 10^{18}\ formula - units}{6.022\times 10^{23}\ formula - units/mol}=5.26\times 10^{-6}\ mol$.

Step2: Find molar mass of $NaIO_3$

The molar mass of $NaIO_3$ is $M_{NaIO_3}=197.89\ g/mol$.

Step3: Calculate mass of $NaIO_3$

$m=n\times M$, where $n = 5.26\times 10^{-6}\ mol$ and $M = 197.89\ g/mol$. So $m=5.26\times 10^{-6}\ mol\times197.89\ g/mol=1.04\times 10^{-3}\ g$