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notsponge [240]
3 years ago
12

Suppose two chemical reactions are linked together in a way that the O2 produced in the first reaction goes on to react complete

ly with Mg to form MgO in the second reaction. Reaction one: 2 KClO3 → 3 O2 + 2 KCl Reaction two: 2 Mg + O2 → 2 MgO If you start with 4 moles of KClO3, how many moles of MgO could eventually form
Chemistry
1 answer:
torisob [31]3 years ago
3 0

Answer:

Numbe of mole of MgO form=12

Explanation:

First calculate the number of mole of O_2 produced from 4 mole of  KCLO_3

Balance the first reaction:

2KClO_3 \rightarrow  3 O_2 + 2 KCl

from the above balanced reaction it is clearly that,

by unitry method,

2 mole of KCLO_3 produced 3 mole of O_2

1 mole of KCLO_3 produced 1.5 mole of O_2

from 4 mole of KCLO_3  4\times 1.5 mole of O_2 produced

hence we have 6 mole of O_2 and this total oxygen will react with Mg

Balance the second reaction:

2Mg + O_2 \rightarrow 2 MgO

since produced oxygen in first reaction is reacted completely with Mg means Mg is given in excess quantity,

From the second balanced reacion it is clearly that,

1 mole of O_2 produced 2 Mole of MgO

hence 6 mole of O_2 will produce 12 mole of MgO

Numbe of mole of MgO form=12

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂     →      CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

                       C             :         CO₂

                        1             :          1

                      1.2           :          1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2  mol × 44 g/mol

Mass = 52.8 g

8 0
3 years ago
One molecule of a compound weighs 2.03x10-22 g. what is the molar mass of that compound?
stealth61 [152]
<span>The molar mass of the compound is 122 g. </span>
6 0
3 years ago
Read 2 more answers
How many moles are in 60.05 g of carbon
Brrunno [24]

Answer:

moles =60.05/12

moles=5.0041

moles=5(approx)

5 0
3 years ago
Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1
Drupady [299]

Answer :  The value of second ionization energy of Ca is 1010 kJ.

Explanation :  

The formation of calcium oxide is,

Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)

\Delta H_f^o = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of CaO:

(1) Conversion of solid calcium into gaseous calcium atoms.

Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)

\Delta H_s = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)

\Delta H_I_1 = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)

\Delta H_I_2 = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

O_2(g)\overset{\Delta H_D}\rightarrow OI(g)

\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)

\Delta H_D = dissociation energy of oxygen = \frac{498}{2}=249kJ

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)

\Delta H_E_1 = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)

\Delta H_E_2 = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)

\Delta H_L = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L

Now put all the given values in this equation, we get:

-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)

\Delta H_I_2=1010kJ

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0
3 years ago
Silver has two naturally isotopes and has an atomic mass of 107.868 amu. One isotope is Ag-109 isotope (108.905 amu) and has a n
Triss [41]

Answer:  106.905

Explanation:  If there are only 2 isotopes, and 1 of them is 48.16%, the second must, by default, be (100 - 48.16%) = 51.84%  The final, averaged, atomic mass is 107.868.  This is made up of each isotope's atomic mass times the percentage of that isotope in the total sample.  The weighted value of the known isotope (109) plus that of the unknown must come to the observed value of 107.868 amu.  (107.868 - 52.45 = 55.42).  Divide that by the % for that isotope (55.42/0.5184) = 106.90 amu for the second isotope.

<u>Atomic Mass</u>  <u>% of Sample</u> <u>Weighted Value</u>

    108.905         48.16%              52.45

          X               51.84%              <u>55.42</u>

                                                     107.87

      X = (55.42/0.5184) = 106.90 amu

5 0
3 years ago
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