Data:
4.24 Kg converting (grams) → 4240 g
Molar Mass of [tex]Na_{2} CO_{3}[/tex]
Na = 2*23 = 46 amu
C = 1*12 = 12 amu
O = 3*16 = 48 amu
-------------------------
Molar Mass of [tex]Na_{2} CO_{3}[/tex] = 46 + 12 + 48 = 106 g/mol
If:
106 g → 1 mol
4240 g → y
Solving: Rule of three (directly proportional)
[tex] \frac{106}{4240} = \frac{1}{y} [/tex]
multiply cross
106*y = 4240*1
106y = 4240
[tex]y = \frac{4240}{106} [/tex]
[tex]\boxed{\boxed{y = 40\:moles}}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
B. 40.0 moles
