Answer:
[tex]m_{O_2}=6gO_2[/tex]
Explanation:
Hello,
In this case, given the reaction:
[tex]2S+3O_2\rightarrow 2SO_3[/tex]
Thus, for 4.00 g of sulfur (atomic mass = 32 g/mol), we apply the shown 2:3 mole ratio with oxygen (molar mass = 32 g/mol) in order to compute its requirement:
[tex]m_{O_2}=4.00gS*\frac{1molS}{32gS}*\frac{3molO_2}{2molS}*\frac{32gO_2}{1molO_2}\\ \\m_{O_2}=6gO_2[/tex]
Best regards.
The mass of oxygen would be required is [tex]6gO_2[/tex]
[tex]= 4.00 \times \frac{1molS}{32gS} \times \frac{3molO_2}{2molS} \times \frac{32gO_2}{1molO_2} \\\\= 6gO_2[/tex]
Here for 4.00 g of sulfur (atomic mass = 32 g/mol), we used the shown 2:3 mole ratio with oxygen (molar mass = 32 g/mol).
Learn more: https://brainly.com/question/17961582?referrer=searchResults
