Question

2n₂h₄(g)+n₂o₄(g)→3n₂(g)+4h₂o(g)
when 8.0g of n₂h₄(32gmol⁻¹) and 92g of n₂o₄(92gmol⁻¹) are mixed together and react according
to the equation above, what is the maximum mass of h₂o that can be produced?
a 9.0g
b 18g
c 36g
d 72g

Explanation:

Step1: Calculate the number of moles of reactants

• For \(N_2H_4\): \(n(N_2H_4)=\frac{m(N_2H_4)}{M(N_2H_4)}=\frac{8.0\ g}{32\ g/mol}=0.25\ mol\)
• For \(N_2O_4\): \(n(N_2O_4)=\frac{m(N_2O_4)}{M(N_2O_4)}=\frac{92\ g}{92\ g/mol}=1\ mol\)

Step2: Determine the limiting reactant

From the balanced equation \(2N_2H_4(g)+N_2O_4(g)\to3N_2(g)+4H_2O(g)\), the mole ratio of \(N_2H_4\) to \(N_2O_4\) is \(2:1\).
If \(0.25\ mol\) of \(N_2H_4\) reacts, it needs \(n(N_2O_4)=\frac{0.25\ mol}{2}=0.125\ mol\) of \(N_2O_4\). Since \(0.125\ mol<1\ mol\), \(N_2H_4\) is the limiting reactant.

Step3: Calculate the moles of \(H_2O\) produced

Using the mole ratio of \(N_2H_4\) to \(H_2O\) (\(2:4 = 1:2\)), if \(n(N_2H_4) = 0.25\ mol\), then \(n(H_2O)=2\times0.25\ mol = 0.5\ mol\)

Step4: Calculate the mass of \(H_2O\)

\(m(H_2O)=n(H_2O)\times M(H_2O)=0.5\ mol\times18\ g/mol = 9.0\ g\)

Answer:

A. \(9.0\ g\)