May 7, 2017 · Na_3PO_4 To find out the limiting reactant, you calculate the number of moles of the reactant and the limiting reactant is the one with the least number of moles. The reaction is; 3BaCl_2 + 2Na_3PO_4 -> Ba_3 (PO_4) _2 + 6NaCl No. of moles = (mass)/(Mr) In this case, No. of moles ( Na_3PO_4) = (5.00)/601.93= 0.00831 moles No. of moles of (BaCl_2) = (6.00/208.23)=0.0288 moles This number of ...

1.363 * 10^(24)" Na"^(+)"cations" If by amount of sodium ions you mean how many of them you have in that sample of sodium carbonate, "Na"_2"CO"_3, then all you have to do here is use two conversion factors one to take you from grams to moles one to take you from moles to number of ions Now, before moving forward, notice that one mole of sodium carbonate contains two moles of sodium cations, 2 ...

Apr 7, 2016 · You have 3 and a bit moles of sodium oxide. "Moles "= " Mass"/"Molar mass" = (200*g)/(61.98*g*mol^-1) = ??" moles" The signal advantage of doing calculations this way, i.e. including units in the calculation is that it adds an extra level of v

Nov 23, 2017 · So, pick one reactant and use the mole ratio that exists between the two reactants to see what you can find. 6.7 color(red)(cancel(color(black)("moles Na"))) * "1 mole Cl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles Na")))) = "3.35 moles Cl"_2 This means that in order for all the moles of sodium to react, you must provide 3.35 moles ...

Aug 12, 2016 · Well, the mass of 1 mol of sodium metal is 22.99*g so you tell us. And 8*molxx22.99*g*mol^-1 = ??g. How many individual sodium atoms are there in such a quantity?

Feb 25, 2016 · Use sodium metal's molar mass to determine how many moles you have in that "11.5-g" sample 11.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.50 moles Na" Now, you know that when 0.50 moles of sodium metal react, the reaction produces 0.50 moles of …

Jun 28, 2015 · To determine the number of moles you have in a certain mass, you need to use molar mass.![http://www.oxnotes.com / http://relative-formula-masses-and-molar-volumes-of ...

Jul 11, 2016 · Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? #2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN#

Jun 2, 2018 · How many moles of #Na^+# ions are in 150 mL of a 0.200 M #Na_3PO_4# solution? Chemistry The Mole Concept The Mole. 1 Answer

Sep 1, 2017 · "28.2 mL" The idea here is that the barium cation delivered to the solution by the barium chloride solution will react with the sulfate anions to produce barium sulfate, "BaSO"_4, an insoluble solid that will precipitate out of the solution. The net ionic equation that describes this double-replacement reaction looks like this "Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr Now ...