Question

given the following, find the number of moles:

1. 30 grams of h₃po₄
2. 25 grams of hf
3. 110 grams of nahco₃
4. 1.1 grams of fecl₃

Explanation:

Problem 1: 30 grams of $\ce{H_3PO_4}$

Step 1: Calculate molar mass of $\ce{H_3PO_4}$

Molar mass of $\ce{H}$ = 1 g/mol, $\ce{P}$ = 31 g/mol, $\ce{O}$ = 16 g/mol.
Molar mass of $\ce{H_3PO_4}$ = $3\times1 + 31 + 4\times16$ = $3 + 31 + 64$ = 98 g/mol.

Step 2: Use moles formula ($n = \frac{m}{M}$)

Given mass ($m$) = 30 g, molar mass ($M$) = 98 g/mol.
$n = \frac{30}{98} \approx 0.306$ mol.

Step 1: Calculate molar mass of $\ce{HF}$

Molar mass of $\ce{H}$ = 1 g/mol, $\ce{F}$ = 19 g/mol.
Molar mass of $\ce{HF}$ = $1 + 19$ = 20 g/mol.

Step 2: Use moles formula ($n = \frac{m}{M}$)

Given mass ($m$) = 25 g, molar mass ($M$) = 20 g/mol.
$n = \frac{25}{20} = 1.25$ mol.

Step 1: Calculate molar mass of $\ce{NaHCO_3}$

Molar mass of $\ce{Na}$ = 23 g/mol, $\ce{H}$ = 1 g/mol, $\ce{C}$ = 12 g/mol, $\ce{O}$ = 16 g/mol.
Molar mass of $\ce{NaHCO_3}$ = $23 + 1 + 12 + 3\times16$ = $23 + 1 + 12 + 48$ = 84 g/mol.

Step 2: Use moles formula ($n = \frac{m}{M}$)

Given mass ($m$) = 110 g, molar mass ($M$) = 84 g/mol.
$n = \frac{110}{84} \approx 1.31$ mol.

Answer:

$\approx 0.31$ mol (or 0.306 mol)

Problem 2: 25 grams of $\ce{HF}$