Question

when the unstable compound ni3(s) decomposes, it produces nitrogen and iodine gases according to the reaction 2ni3(s)→n2(g)+3i2(g). suppose 0.406 grams of ni3 decomposes. calculate the total number of moles of n2(g) and i2(g) produced. hint 0.00461 total moles incorrect. remember that n = m/m. attempts: 1 of 5 used save for later submit answer part 1 current attempt in progress

Explanation:

Step1: Determine molar - mass of NI3

The molar - mass of NI3: N has a molar - mass of approximately 14.01 g/mol and I has a molar - mass of approximately 126.90 g/mol. So, \(M_{NI_3}=14.01 + 3\times126.90=14.01 + 380.7=394.71\ g/mol\).

Step2: Calculate moles of NI3

Given the mass of NI3 \(m = 0.406\ g\), using the formula \(n=\frac{m}{M}\), the number of moles of NI3, \(n_{NI_3}=\frac{0.406\ g}{394.71\ g/mol}\approx0.00103\ mol\).

Step3: Use stoichiometry

From the balanced chemical equation \(2NI_3(s)
ightarrow N_2(g)+3I_2(g)\), the mole ratio of \(NI_3\) to \(N_2\) is \(2:1\) and the mole ratio of \(NI_3\) to \(I_2\) is \(2:3\).
The number of moles of \(N_2\) produced, \(n_{N_2}=\frac{1}{2}n_{NI_3}=\frac{1}{2}\times0.00103\ mol = 0.000515\ mol\).
The number of moles of \(I_2\) produced, \(n_{I_2}=\frac{3}{2}n_{NI_3}=\frac{3}{2}\times0.00103\ mol=0.001545\ mol\).

Step4: Calculate total moles

The total number of moles of \(N_2\) and \(I_2\) produced is \(n_{total}=n_{N_2}+n_{I_2}=0.000515\ mol + 0.001545\ mol=0.00206\ mol\).

Answer:

0.00206