If a student reacted 56.4 grams of stannic nitrate with 32.4 grams of sodium pho

Question

If a student reacted 56.4 grams of stannic nitrate with 32.4 grams of sodium phosphate which reagent is in excess and by how many grams? A student heat a piece of pure unknown metal in a stream of nitrogen gas to form a metal nitride. From the following data determine if he formed sodium nitride, magnesium nitride or aluminum nitride. Your work must match your answer. Wt. of crucible plus lid 24.86 grams Wt. of crucible plus lid plus metal 26.92 grams Wt of crucible plus lid plus product 27.71 grams How many molecules of solid arsenic (V) carbonate must be decomposed to produce 63.4 grams of gas? How many total atoms are in this number of molecules of stannic carbonate?

Explanation / Answer

Answer (1)

First of all we will calculate mole of each reactant, so

Mole of Stannic Nitrate = (weight /Molecular weight)

= (56.4 g / 366 g/mol)

Mole of Stannic Nitrate= 0.154mol

Mole of sodium phosphate = (weight /Molecular weight)

= (32.4 g /120 g/mol)

Mole of sodium phosphate = 0.27 mol

So here the Stannic Nitrate is limiting component and sodium phosphate is excess component by

= 0.27 - 0.154 = 0.116 mol

So sodium phosphate is excess by = 0.116 x 120 = 18.21 gram

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.