A (6.12x10^1) .00 g sample of strontium reacts with an excess of nitrogen to form strontium nitride. What mass of strontium nitride forms?

Question

Olivia Anton · Accepted Answer

Approx. #0.7*g#....

We need a stoichiometric reaction to represent the reaction... #3Sr(s) + N_2(g) rarr Sr_3N_2(s)# Alternatively.... #Sr(s) + 1/3N_2(g)rarr1/3Sr_3N_2# This is in fact a redox equation....and dinitrogen is FORMALLY reduced to give two equiv of #N^(3-)#, i.e. nitride anion... #"Moles of metal"=(0.612*g)/(87.621*g*mol^-1)=6.98xx10^-3*mol# And given the stoichiometry, we can make one third an equiv of strontium nitride....i.e. a mass of .... #1/3xx6.98xx10^-3*molxx290.87*g*mol^-1=0.677*g# I think this nitriding reaction even occurs if you burn strontium in air....you get some of the oxide, as well as some of the nitride....

Cameron Andrews · Answer

undefined To find the mass of strontium nitride formed, you need to determine the molar mass of strontium (Sr) and nitrogen (N), and then use stoichiometry to calculate the mass of strontium nitride formed based on the balanced chemical equation for the reaction between strontium and nitrogen.

1. Calculate the molar mass of Sr:
   Molar mass of Sr = 87.62 g/mol

2. Since the sample is 6.12 × 10^1 g, convert it to moles of Sr:
   Moles of Sr = mass of sample / molar mass of Sr

3. Write the balanced chemical equation for the reaction between strontium and nitrogen:
   3Sr + N2 → Sr3N2

4. According to the balanced chemical equation, 3 moles of strontium react with 1 mole of nitrogen to form 1 mole of strontium nitride (Sr3N2).

5. Calculate the moles of Sr3N2 formed using stoichiometry:
   Moles of Sr3N2 = Moles of Sr (since strontium is the limiting reactant)

6. Convert moles of Sr3N2 to grams using its molar mass:
   Mass of Sr3N2 = Moles of Sr3N2 × molar mass of Sr3N2

By following these steps, you can find the mass of strontium nitride formed when 6.12 × 10^1 g of strontium reacts with an excess of nitrogen.