Question

1. calculate the number of moles in 5.56 x 10^24 atoms of sodium sulfate.
2. calculate the number of moles present in 14.6 grams of ammonium phosphate.
3. how many moles are in 49.7 l of oxygen gas at stp?

Explanation:

1.

Step1: Identify given values

$N = 5.56\times 10^{24}$ atoms (number of atoms of sodium - sulfate), $N_A=6.02\times 10^{23}\text{ atoms/mol}$ (Avogadro's number)

Step2: Apply mole - atom relationship formula

The formula to find the number of moles $n$ is $n=\frac{N}{N_A}$.
Substitute the values: $n=\frac{5.56\times 10^{24}}{6.02\times 10^{23}}$

Step3: Calculate the number of moles

$n=\frac{5.56\times 10^{24}}{6.02\times 10^{23}}=\frac{5.56}{6.02}\times10^{24 - 23}\approx0.9236\times10 = 9.236\approx9.2\text{ mol}$

2.

The molar mass of ammonium phosphate $(NH_4)_3PO_4$:

• $M_N = 14\text{ g/mol}$, $M_H=1\text{ g/mol}$, $M_P = 31\text{ g/mol}$, $M_O = 16\text{ g/mol}$
• $M_{(NH_4)_3PO_4}=3\times(14 + 4\times1)+31+4\times16$
• $=3\times(14 + 4)+31 + 64$
• $=3\times18+31 + 64$
• $=54+31 + 64$
• $=149\text{ g/mol}$

Step1: Identify given values

$m = 14.6\text{ g}$ (mass of ammonium phosphate), $M = 149\text{ g/mol}$ (molar mass of ammonium phosphate)

Step2: Apply mass - mole relationship formula

The formula to find the number of moles $n$ is $n=\frac{m}{M}$
Substitute the values: $n=\frac{14.6}{149}\approx0.098\text{ mol}$

3.

At STP (Standard Temperature and Pressure, $T = 273\text{ K}$, $P = 1\text{ atm}$), the molar volume of a gas $V_m=22.4\text{ L/mol}$

Step1: Identify given values

$V = 49.7\text{ L}$ (volume of oxygen gas), $V_m = 22.4\text{ L/mol}$ (molar volume at STP)

Step2: Apply volume - mole relationship formula

The formula to find the number of moles $n$ is $n=\frac{V}{V_m}$
Substitute the values: $n=\frac{49.7}{22.4}\approx2.22\text{ mol}$

Answer:

1. $9.2\text{ mol}$
2. $0.098\text{ mol}$
3. $2.22\text{ mol}$