Question

calculate the ratio of molecules present in 6.6g of co2 and 3.2g of so2( atomic mass of C= 12u , s=32u, o=16u)​

Answer 1

Answer:

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Explanation:

Answer 2

Answer:-

Required ratio is 3 : 1 .

Explanation:-

For 6.6g CO₂ :-

• Molar mass of Carbon (C) = 12g/mol

• Molar mass of Oxygen (O) = 16g/mol

Hence, molar mass of CO₂ :-

= 12 + 16×2

= 44g/mol

Number of mole in 6.6g CO₂ :-

= Given Mass/Molar mass

= 6.6/44

= 0.15 mole

Number of molecules :-

= No of mole × Avogadro Number

= 0.15×6.022×10²³

= 0.9033×10²³

= 9.033×10²² molecules

For 3.2g SO₂ :-

• Molar mass of Sulphur (S) = 32g/mol

Hence, molar mass of SO₂ :-

= 32+16×2

= 32 + 32

= 64g/mol

Number of mole in 3.2g SO₂ :-

= 3.2/64

= 0.05 mole

Number of molecules :-

= No of mole × Avogadro Number

= 0.05×6.022×10²³

= 0.3011×10²³

= 3.011×10²² molecules

_______________________________

Thus, ratio of molecules present in 6.6g of CO₂ and 3.2g SO₂ :-

= 9.033×10²²/3.011×10²²

= 9.033/3.011

= 3/1

= 3 : 1