Question

the appropriate reduction of sodium carbonate per month is 424 into 106 gram while that of methyl alcohol is 320 into 106. which of the following is produced more in terms of number of moles​

Answer 1

Answer:

Molar weight of

{ Na }_{ 2 }C{ O }_{ 3 }=2\times 23+12+3\times 16=106

Na

2

CO

3

=2×23+12+3×16=106

Thus

\cfrac { 424\times { 10 }^{ 6 } }{ 106 } =4\times { 10 }^{ 6 }

106

424×10

6

=4×10

6

moles of

{ Na }_{ 2 }C{ O }_{ 3 }

Na

2

CO

3

produced

Also molar weight of

C{ H }_{ 3 }OH=12+3\times 1+16+1=32

CH

3

OH=12+3×1+16+1=32

Thus

\cfrac { 320\times { 10 }^{ 6 } }{ 32 } ={ 10 }^{ 5 }

32

320×10

6

=10

5

moles of

C{ H }_{ 3 }OH

CH

3

OH

produced. Hence clearly

C{ H }_{ 3 }OH

CH

3

OH

is produced more in terms of moles.

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Answer 2

$\huge\star\mathfrak\purple{{HEY MATE....!!}}$

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{w}\blue{E}\pink{R}}}}$.

• Moles of sodium carbonate produced per month

=$424 × {10}^6/106 = 4× {10}^6$

• Moles of Methyl alcohol produced per month = $320×{10}^6/32={10}^7$

• So Methyl alcohol produced in terms of moles is more.

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