Question

1. how many moles are present in 120 g of caco₃?
2. calculate the mass of 1.2 moles of kbr.

part b. word problems

1. a chemist has 5.00 g of ch₄. how many moles of methane are present?
2. if a reaction requires 0.50 mole of al, what mass of aluminum is needed?
3. a student prepares 36 g of h₂so₄. calculate the number of moles.
4. how many grams of mg(oh)₂ are in 0.25 moles?
5. a chemist has 80 g of na₂so₄. find the number of moles.

Explanation:

Question 4

Step1: Find molar mass of $CaCO_3$

Molar mass of $Ca$ = 40 g/mol, $C$ = 12 g/mol, $O$ = 16 g/mol.
Molar mass of $CaCO_3$ = $40 + 12 + (3\times16)$ = $40 + 12 + 48$ = 100 g/mol.

Step2: Use moles formula ($n = \frac{m}{M}$)

Given $m = 120$ g, $M = 100$ g/mol.
$n = \frac{120}{100}$ = 1.2 mol.

Step1: Find molar mass of $KBr$

Molar mass of $K$ = 39 g/mol, $Br$ = 80 g/mol.
Molar mass of $KBr$ = $39 + 80$ = 119 g/mol.

Step2: Use mass formula ($m = n\times M$)

Given $n = 1.2$ mol, $M = 119$ g/mol.
$m = 1.2\times119$ = 142.8 g.

Step1: Find molar mass of $CH_4$

Molar mass of $C$ = 12 g/mol, $H$ = 1 g/mol.
Molar mass of $CH_4$ = $12 + (4\times1)$ = 16 g/mol.

Step2: Use moles formula ($n = \frac{m}{M}$)

Given $m = 5.00$ g, $M = 16$ g/mol.
$n = \frac{5.00}{16}$ = 0.3125 mol.

Answer:

1.2 moles

Question 5