Question

Please answer these!!!

Answer 1

1.b. 2.5 mol

2.c.82.8%

3.d. synthesis

4.c . 2:2

5.d. 5 moles

6.c.5.02

7.d.3:1

8.d.2.68 moles

Step-by-step explanation:

1.

`2RbNO_(3)\rightarrow 2RbNO_(2)+1O_(2)`

Here , 2 mole of RbNO3 will produce, 1 mole of O2

1 mole of RbNO3 will produce =

`(1)/(2)` mole of O2

5 mole of RbNO3 will produce =

`(1)/(2)* 5`

= 2.5 mole of O2

2 .

Molar mass of HCl = 1 + 35.45 = 36.45 gram

Molar mass of CO2 = 12 +32 = 44 gram

1 mole of the substance = molar mass

So , 1 mole of HCl = 36.45 gram

2 mole of HCl = 2 x 36.45 =72.9 gram

1 mole of CO2 = 44 gram

The given equation is :

`CaCO_(3)+2HCl\rightarrow CO_(2)+H_(2)O`

Here CaCO3 is the excess reagent and HCl is the limiting reagent

2 mole of HCl will produce = 1 mole of CO2

72.9 gram of HCl= 44 gram of CO2

so, 1 gram of HCl =

`(44)/(72.9)` gram of CO2

Here we need to calculate the mass of CO2 in 200 gram of HCl.

So, 200 gram of HCl =

`(44)/(72.9)* 200` gram CO2

=120.71 gram CO2

this is the theoritical yield of CO2 = 120.71 gram

Experimental yield = 100 gram (given)

The percent yield is calculated by:

`=(Experimental)/(Theoritical)* 100`

`=(100)/(120.71)100`

= 82.84%

3.Synthesis Reaction : Those reactions in which two or more substances combine to give a single product.

Here NH3 and HCl are combined and giving only one product = NH4Cl

4.

`2H_(2)+1O_(2)\rightarrow 2H_(2)O`

Hydrogen = H2

Water = H2O

Here Ratio of H2 to H2O is 2:2

5.The balanced equation is :

`1Ca(OH)_(2)+2HCl\rightarrow 1CaCl_(2)+2H_(2)O`

The number of atoms on left side should be equal to the right side:

Ca = 1

H = 4

O = 2

Cl = 2

Here, 2 moles of water(H2O) = 2 moles of HCl

So, 5 moles of water will give = 5 moles of HCl

6.

`2Ag(NO)_(3)+BaCl_(2)\rightarrow 2AgCl_(2)+BaNO_(3)`

Molar mass of AgNO3 = 169.87 g/mol

1 mol of AgNO3 = 169.87 gram

Molar mass of AgCl = 143.32 g/mol

1 mole of AgCl = 143.32 grams

According to the given equation ,

2 mole of AgNO3 will give = 2 mole of AgCl

1 mole of AgNO3 = 1 mole of AgCl

169.87 gram of AgNO3 = 143.32 gram of AgCL

1 gram of AgNO3 =

`(143.32)/(169.87)`

5.95 gram of AgNO3 =

`(143.32)/(169.87)* 5.95`

= 5.02 grams

7. The mole ratio of Hydrogen (H2) to Nitrogen (N2):

`N_(2)+3H_(2)\rightarrow 2NH_(3)`

H2 = 3 mole

N2 = 1 mole

[H2]:[N2] = 3:1

8. The moles in a substance can be calculated by using :

`moles=(Mass)/(Molar\ mass)`

Mass of N2 = 75 grams (Given)

Molar mass = 28 gram

`moles=(75)/(28)`

Moles = 2.678 moles