Tin(III) Acetate and Magnesium Hydroxide react in a double replacement reaction.

Question

Tin(III) Acetate and Magnesium Hydroxide react in a double replacement reaction. How many grams of magnesium hydroxide are needed to produce 3.97 time 10^22 partocles of tin(III) hydroxide? 5.74 g mg If 26 moles of potassium chloride are formed when potassium and chlorine are combined, how many grams of potassium a required to react with chlorine? 1,014 gk If 15.9 moles of manganese (IV) oxide react with Alumimum, how many particles of Aluminum oxide a formed? 6.38 times 10^24 map Magnesium Chloride reacts with Tim (IV) Sulfide. How many particles of magnesium sulphide will be formed if 260.523 grams of Tin (IV) Chloride are also formed? 1.21 times 10^24 map Mg Potassium Chlorate decomposes to form potassium Chloride and oxygen. If 4.214 times 10^24 particles of potassium Chloride decomposes, how many moles of oxygen are formed? 10.5 mol O_2

Explanation / Answer

Answer – 9) We are given particles of Sn(OH)3 = 3.97*1022

We know reaction –

2 Sn(OAc)3 + 3 Mg(OH)2 ----> 2 Sn(OH)3 + 3 Mg(OAc)2

Now first we need to calculate the moles from particles

We know

6.023*1023 particles = 1 moles

3.97*1022 particles = ?

= 0.0659 moles

From the balanced reaction –

2 moles of Sn(OH)3 = 3 moles of Mg(OH)2

0.0659 moles of Sn(OH)3 = ?

= 0.0989 moles of Mg(OH)2

Mass of Mg(OH)2 = 0.0989 moles * 58.32 g/mole

                              = 5.77 g

10) We are given, moles of KCl = 26 moles

Reaction – 2 KCl + Cl2 ----> 2 KCl

We know

2 moles of KCl = 2 moles of K

So, 26 moles of KCl = ?

= 26 moles of K

Now mass of K

Mass of K = 52 moles * 39.098 g/mol

                   = 1016.5 g

11) We are given, moles of MnO2 = 15.9 moles

Reaction –

2 MnO2 + 4 Al ----> 2 Al2O3 + 2 Mn4+

From the above balanced reaction –

2 moles of MnO2 = 2 moles of Al2O3

15.9 moles of MnO2 = ?

= 15.9 moles of Al2O3

Now need to convert the moles to particles

We know

1 moles = 6.023 *1023 particles

So, 15.9 moles of Al2O3 = ?

= 9.58*1024 particles of Al2O3

12) We are given, mass of SnCl4 = 260.523 g

Reaction – 2 MgCl2 + SnS2 ----> SnCl4 + 2 MgS

Now we need to calculate the moles of SnCl4

Moles of SnCl4 = 260.523 g / 260.5 g.mol-1

                          = 1.0 moles

From the balanced equation –

1 moles of SnCl4 = 2 moles of MgS

So, 1.0 moles of SnCl4 = ?

= 2 moles of MgS

Now need to convert the moles to particles

We know

1 moles = 6.023 *1023 particles

So, 2.0 moles of MgS = ?

= 1.20*1024 particles of MgS

13) We are given the , particles of KCl = 4.214*1024 particles

We know reaction – 2 KClO3 -----> 2 KCl + 3 O2

Now need to convert the moles to particles

We know

6.023 *1023 particles = 1 moles

So, 4.214*1024 particles = ?

= 6.99 moles of KCl

We know,

2 moles of KCl = 3 moles of O2

6.99 moles of KCl = ?

= 10.5 moles O2

So, 10.5 moles of O2 formed.

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