If 102 grams of Sn(NO2)4 is reacted with 80.2 grams of Pt3N4, and yields 40.2 grams of Sn3N4. What is the limiting reagent in the reaction and what is the percent theoretical yield?

Explanation:

The reaction that takes place is

Now we write down the molar mass of Sn(NO₂)₄, Pt₃N₄ and Sn₃N₄:

In order to determine the limiting reagent, we calculate the moles of Sn(NO₂)₄ and Pt₃N₄:

From the balanced reaction we know that 1 mole of Pt₃N₄ would react completely with 3 moles of Sn(NO₂)₄, so 0.319 moles of Pt₃N₄ would require (0.319*3) 0.957 moles of Sn(NO₂)₄ to completely react.

There are not so many Sn(NO₂)₄ moles so Sn(NO₂)₄ is the limiting reagent.

To calculate the yield, first we calculate the moles of Sn₃N₄ produced:

And we compare it with the amount that the limiting reagent (Sn(NO₂)₄) could produce:

So the percent theoretical yield is:

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

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explanation: