Na2CO3(g) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
How many moles of sodium trioxocarbonate (IV) are there in a 25 cm3 solution which required 10cm3 of 0.05 mol dm-3 hydrochloric acid solution to neutralize it?
- 1.000 mole
- 0.100 mole
-
0.010 mole`
- 0.111 mole
The correct answer is: C
Explanation
\(\frac{C_A V_A}{C_BV_B} = \frac{a}{b}\)
\(\frac{0.05 \times 10}{C_B \times 25} = \frac{2}{1}\)
50CB = 0.5
CB = \(\frac{0.5}{50} = \frac{1}{100}\)
= 0.01