# How many particle or molecules of Na_2CO_3 are produced when 10g of NaOH is used? 22.3 x 10^24 particles of NaOH are needed to produce how many liters of H_2O? How many particles/molecules of NaOH are from 30 gram of NaOH?

## $2 N a O H \left(s\right) + C {O}_{2} \left(g\right) \to N {a}_{2} C {O}_{3} \left(s\right) + {H}_{2} O \left(l\right)$

Aug 2, 2017

a) $7.5 \times {10}^{22}$ molecules of $N {a}_{2} C {O}_{3}$ are produced
b)cannot be answered as stoichiometry deals with gases not liquids and water is liquid
c)There are $4.5 \times {10}^{23}$ molecules of $N a O H$ in $30 g$

#### Explanation:

$2 N a O H \left(s\right) + C {O}_{2} \left(g\right) \to N {a}_{2} C {O}_{3} \left(s\right) + {H}_{2} O \left(l\right)$

Molar mass of $N a O H = 40$

$2 \times 40 g$ of $N a O H$ produces $6.02 \times {10}^{23}$ molecules of $N {a}_{2} C {O}_{3}$
$\therefore$ $10 g$ of $N a O H$ produces $\frac{6.02 \times {10}^{23} \times 10}{2 \times 40}$ molecules of $N {a}_{2} C {O}_{3}$
$= 7.5 \times {10}^{22}$ molecules of $N {a}_{2} C {O}_{3}$

$40 g$ of $N a O H$ contains $6 \times {10}^{23}$ molecules
$\therefore$ $30 g$ of $N a O H$ contains $\frac{30 \times 6 \times {10}^{23}}{40}$ molecules
$= 4.5 \times {10}^{23}$ molecules