Question

stoichiometry help! please let me know if more information is needed as i do have some.

NaNO2ag5+ HSOŞNHącag+NaHSO4cags + N2g9+ H20.00 v 10mL (0.011) 8.00mL (0.008L). molarity of HSONH2= 0.803 malth molarity of NaNO₂ = 0. 250 mol/L | how many moles of N2(g) are produced?

Answer 1

Answer:

In the above reaction, 10 mL of NaNO₂ contains

0.250 moles = 1000 mL (from concentration given)

Moles of NaNO₂ = 2.50 x 10^-3 moles

And the moles of sulfamic acid present in 8 mL solution

0.803 moles = 1000 mL (from concentration given)

Moles of HSO₃NH₂ = 6.424 x 10^-3 moles

Here, NaNO₂ is the limiting reagent. Hence, the number of moles N₂ form will be equal to 2.50 x 10^-3.

As in the above reaction, 1 mole of NaNO₂ is required by HSO₃NH₂, which will produce the 1 mole of N₂.

Hence, moles of N₂ =  2.50 x 10^-3 moles

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