Consider the double replacement between calcium sulfate (CaSO4) and Sodium Iodid
ID: 901356 • Letter: C
Question
Consider the double replacement between calcium sulfate (CaSO4) and Sodium Iodide (Nal). If 34.7g of Calcium sulfate and 58.3 g of sodium Iodide are placed in a reaction vessel , how many grams of each product are produced.
A) write the balance equation
B)If you use up all the calcium sulfate, how many moles of calcium iodide will get produced(convert grams of calcium sulfate to moles using the molar mass; then use the mole of ratio from balanced chemical equation to figure out moles of product
C)If you use up all of the sodium iodide, how many moles of calcium iodide will get produced? (convert grams of sodium iodide to moles using the molar mass; then use mole ratio from balanced chemical equation to figure out moles of product)
D) Which reactant makes the smallest amount of product ? That will be the reactant that gets used up first and is defined as your limiting reacting
E) Now that you know which reactant get used up, go back to part B or C and convert moles of calcium iodide to grams using the molar mass of calcium iodide.
Explanation / Answer
A) CaSO4 + 2NaI --> CaI2 + Na2SO4
B) Number of moles of CaSO4,n = m/M = 34.7/136 = 0.255 moles
1 mole of CaSO4 reacts 2 moles of NaI as per equn
0.255 moles of CaSO4 reacts ?
= 0.255 * 2 /1 = 0.51 moles of CaSO4
C) Number of moles of NaI = 58.3 /150 = 0.389
2 moles of NaI reacts 1 mole CaSO4
0.389 moles of NaI reacts ?
= 0.389 * 1/2 = 0.1945 moles
D) NaI is limiting reagent . 0.1945 moles of Na2SO4 and CaI2 produced.
E) Amount of Na2SO4, m = n*M = 0.1945 * 142.04 = 27.6g
Amount of CaI2 = 0.1945 * 294 = 57.18g
So. small amount of product is Na2SO4