1. Lead (II) nitrate and sodium iodide react to form sodium nitrate and lead (II

Question

1. Lead (II) nitrate and sodium iodide react to form sodium nitrate and lead (II) iodide.

Pb(NO3)2 + NaI --> NaNO3 + PbI2

A)    Balance the equation.

B)    Name the type of reaction.

C)    How many moles of sodium iodide are needed to produce 225g PbI2?

D)    If 234g NaI are used, how many grams of PbI2 are produced?

E)    How many molecules of NaNO3 are produced from 637g Pb(NO3)2?

2. A 26.0mL sample of phosphoric acid (H3PO4) is neutralized by 32.7mL of 2.46M NaOH.

A)    Write the products and balance the equation:

NaOH + H3PO4 -->

B)    What is the molarity of the H3PO4 solution?

Explanation / Answer

(1) (A) Pb(NO3)2 + 2 NaI => 2 NaNO3 + PbI2


(B) This is a precipitation (double-displacement/metathesis) reaction


(C) Moles of PbI2 = mass/molar mass of PbI2

= 225/461.02 = 0.488 mol


Moles of NaI = 2 x moles of PbI2

= 2 x 0.488 = 0.976 mol


(D) Moles of NaI = mass/molar mass of NaI

= 234/149.89 = 1.5611 mol


Moles of PbI2 = 1/2 x moles of NaI

= 1/2 x 1.5611 = 0.7806 mol


Mass of PbI2 = moles x molar mass of PbI2

= 0.7806 x 461.02 = 360 g


(E) Moles of Pb(NO3)2 = mass/molar mass of Pb(NO3)2

= 637/331.22 = 1.923 mol


Moles of NaNO3 = 2 x moles of Pb(NO3)2

= 2 x 1.923 = 3.846 mol


Number of NaNO3 = moles x Avogadros number

= 3.846 x 6.022 x 10^23 = 2.32 x 10^24 molecules


(2) (A) H3PO4 + 3 NaOH => 3 H2O + Na3PO4


(B) Moles of NaOH = volume x concentration of NaOH

= 32.7/1000 x 2.46 = 0.080442 mol


Moles of H3PO4 = 1/3 x moles of NaOH

= 1/3 x 0.080442 = 0.026814 mol


Concentration of H3PO4 = moles/volume of H3PO4

= 0.026814/0.0260 = 1.03 M

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