Question

General Chem: questions 1-10

1. 9.73 g of lead(IV) chloride contains enough Cl- ions to make ____ g of magnesium chloride.

a) 5.31 g    b) 0.188 g     c) 17.8 g     d) 1.32 g     e) 0.0557 g

2. What are the spectator ions when Co(OH)3 reacts with H2SO4?

a) none     b) H+ and OH-     c) Co+3 and SO4-2 d) SO4-2 e) Co+3

3. When the molecular reaction between sodium silicate and copper(II) nitrite is balanced correctly the stoichiometric coefficient for sodium nitrite is _____.

a) 1     b) 2     c) 4     d) 8     e) 3

4. How many FREE ions are there on the product side of the total ionic equation that results from the reaction between bismuth(III) hypochlorite and acetic acid.

a) 0     b) 10     c) 6     d) 4     e) 8

5. What is the net ionic equation for the reaction between tin(IV) sulfide and nitric acid?

            a) SnS2 + 4 H+ ---> 2 H2S + Sn+4

            b) S-2 + 2 H+ ---> H2S

            c) SnS2 + 4 HNO3 ----> 2 H2S + Sn(NO3)4

            d) Sn+4 + 4 NO3- ----> Sn(NO3)4

            e) SnS2 + 4 NO3- ----> 2 S-2 + Sn(NO3)4

6. Which of the combinations below will produce an insoluble salt?

a) Ba(OH)2 + HCl b) MnSO4 + Pb(NO2)2     c) H2SO4 + AlBr3

d) SrS + HClO     e) MgCl2 + Au(ClO4)3

7. I2     + Ti+4 ---->    Ti +     IO3-    

When the full reaction above is balanced, what is the stoichiometric coefficient for Ti?

a) 4     b) 10         c) 5     d) 3     e) 1

8. For the reaction in question 7, which species is the reducing agent

a) Ti     b) Ti+4     c) I2     d) IO3-     e) H2O

9. The combustion of 28.8 g of NH3 consumes exactly _____ g of O2.       

  4 NH3 + 7 O2 ----> 4 NO2 + 6 H2O

a) 94.9     b) 54.12     c) 108     d) 15.3     e) 28.8

10. 3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced?

                        6 Li + N2 ---> 2 Li3N

a) 3.52 g     b) 2.93 g     c) 17.6 g     d) 5.85 g     e) 8.70 g

Answer 1

Question 1. Given that, Mass of PbCl2 = 9.73g Molar mass of PbCl. = 349.01 g/mol Calculate the number of moles of PbCl, as follows: mass Moles of PbCl. =- molar mass 9.73 g =.0279 mol 349.01 g/mol Imol of PbCl2 = 4 mol of Cl ions = 4x.0279 mol = 0.1116 mol of CI lmol of MgCl, = 2 mol of Clions = 2 x.0279 mol = 0.0558 mol of CI Calculate the mass of magnesium chloride that contains enough chloride ions as follows:

Mass of MgCl, = Molar mass xMoles =95.21 g/mol x 0.0558 mol = 5.31g Therefore, the mass of magnesium chloride that contains enough chloride ions is 5.31g and the correct option is (A). Question 2. Spectator ions: These are the ions which are present on reactant as well as on product side but are not involved into the overall reaction. The reaction between Co(OH), and H2SO4 is given as follows: CO(OH), +H,804 → Co2+ +503 + H20 Therefore, the spectator ions present when Co(OH), reacts with H2SO4 are Cost and so and the correct option is (C).

Question 3. The reaction between sodium silicate, Na SiOz, and copper nitrite, Cu(NO2),, is given as follows: Na Sioz+Cu(NO2), → 2NaNO, + Cusioz Sodium silicate Copper nitrite Sodium nitrite Copper silicate Therefore, in the balanced molecular reaction between Na Sio, and Cu(NO2),, the stoichiometric coefficient of sodium nitrite, NaNO,, is 2 and the correct option is (B). Question 4. The molecular reaction between bismuth hypochlorite, Bi (CIO), , and acetic acid, CH,COOH, is given as follows: Bi(CIO), +3CH2COOH → Bi(CH3COO), +3HCIO Bismuth hypochlorite

The total ionic reaction between bismuth hypochlorite, Bi(CIO), , and acetic acid, CH,COOH, is given as follows: Bi (CIO), +3CH,C00+3H* →Bi*: +3CH,C00+3HCIO Therefore, the total number of free ions present on product side of the total ionic reaction are 4 and the correct option is (D). Question 5. The balanced molecular reaction between tin (IV) sulphide and nitric acid is given as follows: SnS, +4HNO3 → 2H S+Sn(NO3). The total ionic reaction between tin (IV) sulphide, SnS,, and nitric acid, HNO3, is given as follows:

Sns, + 4H+ + 4NO; → 2H,S+Sn** + 4NO SnS, +4H* → 2H,S+Sn +

Therefore, the total ionic reaction between tin (IV) sulphide, SnS2, and nitric acid, HNO3, is Sns, + 4H2H,S +Sn" and the correct option is (A). Question 6. The given combinations are, (a). Ba(OH), +HCI (b). MnSO4 + Pb(NO2), (c). H2SO4 +AIBI (d). SIS + HCIO (e). MgCl2 + Au(C104); All the given combinations are acid base combinations which results into the formation of salt. In given combinations, all forms soluble salts except that of the combination (b). the reaction is given as follows: MnSO4 + Pb(NO2), → PbSO4 + Mn (NO2)2 Insoluble salt

Therefore, the combination MnSO4 + Pb(NO2), results into the formation of insoluble salt, PbSO4, and the correct option is (B). Question 7. The given reaction is, 12 + Ti^+ → Ti+10; The above chemical reaction is balanced by separating it into two different half reactions and the complete balanced reaction is given as follows: 21, +57i4+ +12H,0 + 5Ti+4103 +24H* Therefore, when the reaction is balanced, the stoichiometric coefficient for titanium is 5 and the correct option is (C). Question 8.

Reducing agent: A substance that brings about the reduction by oxidised and loss of electrons by itself. For the above reaction in question 7, the reducing agent is I, and the correct option is (C). Question 9. The given reaction is, 4NH, +70, +4NO2 +6H2O Mass of NH, = 28.8 g Calculate the mass of o, consumes in the reaction as follows: Calculate the number of moles of NH, as follows:

mass Moles of NH, = molar mass 28.8 g 17.03 g/mol = 1.69 mol Calculate the number of moles of O, as follows: 4 mol of NH, consumes = 7 mol of O, 7 mol 1.69 mol of NH, consumes = -10, x1.69 mol of O, 4mol = 2.96 mol 1 mol of O, = 32 g 32g 2.96 mol of O, 8 x 2.96 mol 1mol =94.9g

Therefore, the mass of O, consumes in the reaction is 94.9g and the correct option is (A). Question 10. The reaction is given as follows:

6 Li+N, +2L1,N Mass of Li=3.50 g Mass of N, = 3.50 g Calculate the maximum mass of Li N produced during the reaction as follows: Calculate the number of moles of Li and N, as follows: Mass Moles of Lir. Molar mass 3.50 g 6.9 g/mol = 0.51mol Mass Moles of N2= Molar mass 3.50 g 14 g/mol = 0.125 mol

moles of N, reacted = 0.084042 mol moles of LiN produced = 2 x0.084042 mol=0.16808 mol mass of Li N = 0.16808molx 34.83 g/mol = 5.85 g Therefore, the maximum mass of LiN produced during the reaction is 5.85 g and the correct option is (D).