1) How many total atoms are in 0.640g of P2O5? - ________ atoms 2) A hydrated fo

Question

1) How many total atoms are in 0.640g of P2O5?

- ________ atoms

2) A hydrated form of copper sulfate (CuSO4 x H2O) is heated to drive off all the water. if we start with 8.64g of hydrated salt and have 4.83g anhydrous CuSO4 after heating, find the number of water molecules associated with each CuSO4 formula units.

- x= ________

3) When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

CaCO3(s)+2HCl(aq)--------> CaCl2(aq)+H2O(l)+CO2(g)

-How many grams of calcium chloride will be produced when 30.0g of calcium carbonate are combined with 13.0g hydrochloric acid?

_______g CaCl2

- Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

_______g of CaCO3 or HCl

4) Assuming an efficiency of 27.60%, calculate the actual yield of magnesium nitrate formed from 117.7g of magnesium and excess copper(ll) nitrate.

Mg+Cu(NO3)2-------> Mg(NO3)2+Cu

- ________g

5) Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s)+2HCl(aq)---------> ZnCl2(aq)+H2(g)

- ________mL

6) Lead(ll) nitrate and ammonium iodide react to form lead(ll) iodide and ammonium nitrate according to the reaction

Pb(NO3)2(aq)+2NH4I(aq)---------->PbI2(s)+2NH4NO3(aq)

-What volume of a 0.430 M NH2I solution is required to react with 773 mL of a 0.580 M Pb(NO3)2 solution?

______mL

-How many moles of PbI2 are formed from this reaction?

________mol PbI2

7) Consider the neutralization reaction below.

2HNO3(aq)+ Ba(OH)2(aq)-----------> 2H2O(l)+Ba(NO3)2(aq)

- A 0.120-L sample of an unknown HNO3 solution required 40.7 mL of 0.200 M Ba(OH)2 for complete neutralization. What was the concetration of the HNO3 solution?

_______M

Explanation / Answer

1) Total atoms = (mass of P2O5/ molar mass of P2O5) x Avogadro number

= (0.64 g/ 142 g/mol) x 6.023 x 1023 mol-1

= 0.027 x 1023

4)   Assuming an efficiency of 27.60%, calculate the actual yield of magnesium nitrate formed from 117.7g of magnesium and excess copper(ll) nitrate.

Mg + Cu(NO3)2 -------> Mg(NO3)2 + Cu

1 mol = 24 g 1mol = 148 g

117.7g ?

? = (117.7 g / 24 g ) x 148 g Mg(NO3)2

= 725.8 g of Mg(NO3)2

This is for 100 % yield.

But given that yield = 27.6 %

Therefore,

actual mass of Mg(NO3)2 = 27.6 % of 725.8 g = 196 g

6)   Pb(NO3)2(aq)+2NH4I(aq)---------->PbI2(s)+2NH4NO3(aq)

Pb(NO3)2:

M1 = 0.58 M , V1 = 773 mL , n1 = no of moles of Pb(NO3)2 in balanced equation = 1

NH4I:

M2 = 0.43 M , V2 = ? , n2 = no of moles of NH4I in balanced equation = 2

We know that

M1V1/n1 = M2V2/n2

V2 = (M1V1/n1) x (n2/M2)

= (0.58 M x 773 mL/ 1) ( 2 / 0.43 M)

= 2085 mL

Therefore,

volume of NH4I required = 2085 mL

7)

2HNO3(aq)+ Ba(OH)2(aq)-----------> 2H2O(l)+Ba(NO3)2(aq)

Ba(OH)2:

M1 = 0.2 M , V1 = 40.7 mL , n1 = no of moles of Ba(OH)2 in balanced equation = 1

HNO3:

M2 = ? , V2 = 0.12 L = 120 ml , n2 = no of moles of HNO3 in balanced equation = 2

We know that

M1V1/n1 = M2V2/n2

M2 = (M1V1/n1) x (n2/V2)

= (0.2 M x 40.7 mL/ 1) ( 2 / 120 mL)

= 0.135 M

Therefore,

concetration of the HNO3 solution = 0.135 M

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