PRELIMINARY STAGE 2026
Contest 19
Wesley Girls’ High School: 59 points
Nkwatia Presby SHS: 26 points
Simms SHS: 22 points
Contest 20
Labone SHS: 41 points
Notre Dame Sem. SHS: 39 points
T.I. AMASS, Wa: 03 points
Contest 21
Winneba Secondary School: 44 points
Fijai SHS: 42 points
Enyan Denkyira SHTS: 12 points
QUESTION
If 56.0 g of nitrogen gas reacts completely with excess hydrogen, what mass of ammonia gas is formed?
ANSWER: 68.0 g
SOLUTION:
$m = \dfrac{m_{\text{given}} \times M_{\text{product}} \times z}{M_{\text{reactant}}}$
Where:
$m$ = mass of ammonia gas formed (g)
$m_{\text{given}}$ = mass of nitrogen gas ($56.0\text{ g}$)
$M_{\text{reactant}}$ = molar mass of $N_2$ ($28.0\text{ g/mol}$)
$M_{\text{product}}$ = molar mass of $NH_3$ ($17.0\text{ g/mol}$)
$z$ = mole ratio coefficient ($2$ for $N_2 \rightarrow 2NH_3$)
$n_{N_2} = \dfrac{56.0}{28.0}$
$n_{N_2} = 2.00\text{ mol}$
$n_{NH_3} = 2.00 \times 2$
$n_{NH_3} = 4.00\text{ mol}$
$m = 4.00 \times 17.0$
$m = 68.0\text{ g}$
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PRACTICE QUESTIONS
1. QUESTION: If 14.0 g of nitrogen gas reacts completely with excess hydrogen, what mass of ammonia is formed? (N = 14.0, H = 1.00)
ANSWER: 17.0 g or $1.70\times10^{1}$ g
SOLUTION:
$N_2+3H_2\rightarrow2NH_3$
$n(N_2) = \dfrac{14.0}{28.0}$
$n(N_2) = \dfrac{1}{2}$
$n(N_2) = 0.500\text{ mol}$
$n(NH_3) = 2\times0.500$
$n(NH_3) = 2 \times 5\times10^{-1}$
$n(NH_3) = 1.00\text{ mol}$
$m = 1.00\times17.0$
$m = 17.0\text{ g}$
2. QUESTION: If 6.00 g of hydrogen reacts completely with excess nitrogen, what mass of ammonia is formed? (N = 14.0, H = 1.00)
ANSWER: 34.0 g or $3.40\times10^{1}$ g
SOLUTION:
$N_2+3H_2\rightarrow2NH_3$
$n(H_2) = \dfrac{6.00}{2.00}$
$n(H_2) = 3.00\text{ mol}$
$n(NH_3) = \dfrac{2}{3}\times3.00$
$n(NH_3) = 2.00\text{ mol}$
$m = 2.00\times17.0$
$m = 34.0\text{ g}$
3. QUESTION: In the Contact process, 32.0 g of sulfur dioxide is completely converted to sulfur trioxide. What mass of sulfur trioxide is formed? (S = 32.0, O = 16.0)
ANSWER: 40.0 g or $4.00\times10^{1}$ g
SOLUTION:
$2SO_2+O_2\rightarrow2SO_3$
$n(SO_2) = \dfrac{32.0}{64.0}$
$n(SO_2) = \dfrac{1}{2}$
$n(SO_2) = 0.500\text{ mol}$
$n(SO_3) = 0.500\text{ mol}$
$m = 0.500\times80.0$
$m = 5\times10^{-1} \times 80$
$m = 40.0\text{ g}$
4. QUESTION: If 3.00 g of hydrogen reacts completely with excess nitrogen, what mass of ammonia is formed? (N = 14.0, H = 1.00)
ANSWER: 17.0 g or $1.70\times10^{1}$ g
SOLUTION:
$N_2+3H_2\rightarrow2NH_3$
$n(H_2) = \dfrac{3.00}{2.00}$
$n(H_2) = 1.50\text{ mol}$
$n(NH_3) = \dfrac{2}{3}\times1.50$
$n(NH_3) = \dfrac{2 \times 15\times10^{-1}}{3}$
$n(NH_3) = 2 \times 5 \times10^{-1}$
$n(NH_3) = 1.00\text{ mol}$
$m = 1.00\times17.0$
$m = 17.0\text{ g}$
5. QUESTION: If 28.0 g of nitrogen gas reacts completely with excess hydrogen, what mass of ammonia is formed? (N = 14.0, H = 1.00)
ANSWER: 34.0 g or $3.40\times10^{1}$ g
SOLUTION:
$N_2+3H_2\rightarrow2NH_3$
$n(N_2) = \dfrac{28.0}{28.0}$
$n(N_2) = 1.00\text{ mol}$
$n(NH_3) = 2\times1.00$
$n(NH_3) = 2.00\text{ mol}$
$m = 2.00\times17.0$
$m = 34.0\text{ g}$
6. QUESTION: In the Contact process, 64.0 g of sulfur dioxide is completely converted to sulfur trioxide. What mass of sulfur trioxide is formed? (S = 32.0, O = 16.0)
ANSWER: 80.0 g or $8.00\times10^{1}$ g
SOLUTION:
$2SO_2+O_2\rightarrow2SO_3$
$n(SO_2) = \dfrac{64.0}{64.0}$
$n(SO_2) = 1.00\text{ mol}$
$n(SO_3) = 1.00\text{ mol}$
$m = 1.00\times80.0$
$m = 80.0\text{ g}$