I got this whole question wrong, someone please find the right answer for me! (1

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I got this whole question wrong, someone please find the right answer for me! (1 & 2 are wrong)

Consider the following unbalanced reaction: Na_3PO_4(aq) + 3 Ba(NO_3)_2(aq) +6 NaNo_3(aq) Suppose that a solution containing 2.53 grams of sodium phosphate (mm = 163.94) is mixed with a solution containing 5.92 grams of barium nitrate (mm = 261.37). How many grams of barium phosphate (mm 601.92) can be formed? A pair of students in the lab did the precipitated reaction of barium phosphate shown in part 1. After they dried the sample they produced, they found that the product weighted 3.978 g. What was their percent yield for the experiment?

Explanation / Answer

1)

mass of Na3PO4 = 2.53
molar mass of Na3PO4 = 163.94
mol of Na3PO4 = (mass)/(molar mass)
= 2.53/163.94
= 0.02

mass of Ba(NO3)2 = 2.53
molar mass of Ba(NO3)2 = 5.92
mol of Ba(NO3)2 = mass/molar mass
= 5.92/261.37
= 0.02

Balanced chemical equation is:
2Na3PO4 + 3Ba(NO3)2 ---> 1Ba3(PO4)2 + 6NaNO3

2 mol of Na3PO4 reacts with 3mol of Ba(NO3)2
for 0.02 mol of Na3PO40.03mol of Ba(NO3)2 is required
But we have 0.02 mol of Ba(NO3)2

Ba(NO3)2 is limiting reagent
we will use Ba(NO3)2 in further calculation
According to balanced equation
mol of Ba3(PO4)2 formed = (1/3)* moles of Ba(NO3)2
= (1/3)*0.02
= 0.01
mass of Ba3(PO4)2 = number of mol * molar mass
= 0.01*601.92
= 4.54

Answer: 4.54 g

i am allowed to answer only 1 question at a time

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