Answer: 111g S
Explanation:
SO2 you have = 665 g
SO2 molar mass = 64.066 g/mol
S molar mass = 32.065 g/mol
[tex]665g SO2*\frac{1 molSO2}{64.066gSO2} *\frac{1molS}{3molSO2} * \frac{32.065gS}{1molS} = 110.944... ~ 111gS[/tex]
There are 332.5 grams of S (sulfur) in 665 g of SO2 ( sulfur dioxide).
From the information given:
The atomic mass of sulfur = 32 g/mol
The atomic mass of oxygen = 16 g/mol
∴
Since there are 665g of SO2;
If 1 mole of SO2 contains 1 mole of S
It implies that:
(1 × 64)g of SO2 will contain (1 × 32) g of S;
∴
665 g of SO2 will contain :
[tex]\mathbf{= \Big ( \dfrac{1 \times 32}{1 \times 64} \times 665 \Big )}[/tex]
= [tex]\mathbf{\Big 332.5 \ grams \ of \ S \Big}[/tex]
Therefore, we can conclude that there are 332.5 grams of S present in 665 g of SO2.
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