PRELIMINARY STAGE 2026
Contest 40
Konongo Odumase SHS: 45 points
Our Lady of Mt. Carmel Girls’ SHS: 31 points
Juaben SHS: 28 points
Contest 41
St. Monica’s SHS: 40 points
Amenfiman SHS: 37 points
Tema Methodist Day SHS: 15 points
Contest 42
Accra High School: 33 points
Ghana SHS, Tamale: 32 points
Sunyani SHS: 31 points
ROUND 2 - SPEED RACE
QUESTION
What mass of sodium hydrogen carbonate, NaHCO$_3$, is required to completely neutralize 50.0 cm$^3$ of 0.400 mol dm$^{-3}$ hydrochloric acid, HCl?
$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$
ANSWER: $1.68\text{ g} \text{ or } 1.68 \times 10^{-3} \text{kg} \text{ or } 0.00168\text{ kg}$
SOLUTION 1:
$C = 4 \times 10^{-1} \text{ mol/dm}^3$
$V = 50 \times 10^0 \text{ cm}^3$
$M = 84 \times 10^0 \text{ g/mol}$
$m = \dfrac{C \times V \times M}{1000}$
$m = \dfrac{4 \times 10^{-1} \times 50 \times 10^0 \times 84 \times 10^0}{1000}$
$m = \dfrac{16800 \times 10^{-1}}{1000}$
$m = \dfrac{1680}{1000}$
$m = 1.68\text{ g}$
SOLUTION 2:
$C = 4 \times 10^{-1} \text{ mol/dm}^3$
$V = 50 \times 10^0 \text{ cm}^3$
$M = 84 \times 10^0 \text{ g/mol}$
$\text{Shortcut constant: } \dfrac{1}{1000} = 1 \times 10^{-3}$
$m = 4 \times 10^{-1} \times 50 \times 10^0 \times 84 \times 10^0 \times 1 \times 10^{-3}$
$m = 2 \times 84 \times10^{-2}$
$m = 16800 \times 10^{-4}\text{ g}$
$m = 1.68\text{ g}$
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PRACTICE QUESTIONS
1. QUESTION: What mass of sodium hydrogencarbonate, $NaHCO_3$, is required to completely neutralise 25.0 cm$^3$ of 0.200 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 0.420 g or $4.20\times10^{-1}$ g
SOLUTION:
$NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2$
$n(HCl) = 0.200\times\dfrac{25.0}{1000}$
$n(HCl) = \dfrac{25 \times 2\times10^{-1}}{1000}$
$n(HCl) = \dfrac{2}{40} \times10^{-1}$
$n(HCl) = \dfrac{1}{20} \times10^{-1}$
$n(HCl) = 5.00\times10^{-3}\text{ mol}$
$n(NaHCO_3) = 5.00\times10^{-3}$
$n(NaHCO_3) = 5.00\times10^{-3}\text{ mol}$
$m = 5.00\times10^{-3}\times84$
$m = 5\times10^{-3} \times 84$
$m = 0.420\text{ g}$
2. QUESTION: What mass of sodium carbonate, $Na_2CO_3$, is required to completely neutralise 40.0 cm$^3$ of 0.250 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 0.530 g or $5.30\times10^{-1}$ g
SOLUTION:
$Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2$
$n(HCl) = 0.250\times\dfrac{40.0}{1000}$
$n(HCl) = \dfrac{40 \times 25\times10^{-2}}{1000}$
$n(HCl) = \dfrac{25}{25} \times10^{-2}$
$n(HCl) = 1.00\times10^{-2}\text{ mol}$
$n(Na_2CO_3) = \dfrac{1.00\times10^{-2}}{2}$
$n(Na_2CO_3) = \dfrac{10^{-2}}{2}$
$n(Na_2CO_3) = 5.00\times10^{-3}\text{ mol}$
$m = 5.00\times10^{-3}\times106$
$m = 5\times10^{-3} \times 106$
$m = 0.530\text{ g}$
3. QUESTION: What mass of calcium carbonate, $CaCO_3$, is required to completely neutralise 50.0 cm$^3$ of 0.400 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 1.00 g
SOLUTION:
$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$
$n(HCl) = 0.400\times\dfrac{50.0}{1000}$
$n(HCl) = \dfrac{50 \times 4\times10^{-1}}{1000}$
$n(HCl) = \dfrac{4}{20} \times10^{-1}$
$n(HCl) = \dfrac{1}{5} \times10^{-1}$
$n(HCl) = 2.00\times10^{-2}\text{ mol}$
$n(CaCO_3) = \dfrac{2.00\times10^{-2}}{2}$
$n(CaCO_3) = \dfrac{2\times10^{-2}}{2}$
$n(CaCO_3) = 1.00\times10^{-2}\text{ mol}$
$m = 1.00\times10^{-2}\times100$
$m = 10^{-2} \times 100$
$m = 1.00\text{ g}$
4. QUESTION: What mass of sodium hydrogencarbonate, $NaHCO_3$, is required to completely neutralise 20.0 cm$^3$ of 0.500 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 0.840 g or $8.40\times10^{-1}$ g
SOLUTION:
$NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2$
$n(HCl) = 0.500\times\dfrac{20.0}{1000}$
$n(HCl) = \dfrac{20 \times 5\times10^{-1}}{1000}$
$n(HCl) = \dfrac{5}{50} \times10^{-1}$
$n(HCl) = \dfrac{1}{10} \times10^{-1}$
$n(HCl) = 1.00\times10^{-2}\text{ mol}$
$n(NaHCO_3) = 1.00\times10^{-2}$
$n(NaHCO_3) = 1.00\times10^{-2}\text{ mol}$
$m = 1.00\times10^{-2}\times84$
$m = 10^{-2} \times 84$
$m = 0.840\text{ g}$
5. QUESTION: What mass of sodium carbonate, $Na_2CO_3$, is required to completely neutralise 25.0 cm$^3$ of 0.200 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 0.265 g or $2.65\times10^{-1}$ g
SOLUTION:
$Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2$
$n(HCl) = 0.200\times\dfrac{25.0}{1000}$
$n(HCl) = \dfrac{25 \times 2\times10^{-1}}{1000}$
$n(HCl) = \dfrac{2}{40} \times10^{-1}$
$n(HCl) = \dfrac{1}{20} \times10^{-1}$
$n(HCl) = 5.00\times10^{-3}\text{ mol}$
$n(Na_2CO_3) = \dfrac{5.00\times10^{-3}}{2}$
$n(Na_2CO_3) = \dfrac{5\times10^{-3}}{2}$
$n(Na_2CO_3) = 2.50\times10^{-3}\text{ mol}$
$m = 2.50\times10^{-3}\times106$
$m = 25\times10^{-4} \times 106$
$m = 0.265\text{ g}$
6. QUESTION: What mass of calcium carbonate, $CaCO_3$, is required to completely neutralise 30.0 cm$^3$ of 0.100 mol dm$^{-3}$ hydrochloric acid? (Na = 23, Ca = 40, H = 1, C = 12, O = 16)
ANSWER: 0.150 g or $1.50\times10^{-1}$ g
SOLUTION:
$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$
$n(HCl) = 0.100\times\dfrac{30.0}{1000}$
$n(HCl) = \dfrac{30 \times 10^{-1}}{1000}$
$n(HCl) = \dfrac{3}{100} \times10^{-1}$
$n(HCl) = 3.00\times10^{-3}\text{ mol}$
$n(CaCO_3) = \dfrac{3.00\times10^{-3}}{2}$
$n(CaCO_3) = \dfrac{3\times10^{-3}}{2}$
$n(CaCO_3) = 1.50\times10^{-3}\text{ mol}$
$m = 1.50\times10^{-3}\times100$
$m = 15\times10^{-4} \times 100$
$m = 0.150\text{ g}$