Data:
4.24 Kg converting (grams) → 4240 g
Molar Mass of \(Na_{2} CO_{3}\)
Na = 223 = 46 amu
C = 112 = 12 amu
O = 316 = 48 amu
-------------------------
Molar Mass of \(Na_{2} CO_{3}\) = 46 + 12 + 48 = 106 g/mol
If:
106 g → 1 mol
4240 g → y
Solving: Rule of three (directly proportional)
\(\frac{106}{4240} = \frac{1}{y}\)
multiply cross
106y = 4240*1
106y = 4240
\(y = \frac{4240}{106}\)
\(\boxed{\boxed{y = 40\:moles}}\end{array}}\qquad\quad\checkmark\)
Answer:
B. 40.0 moles
