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Nikitich [7]
3 years ago
5

After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the R

un Experiment button at the end or by clicking the Overview tab and returning to the Experiment tab to select Run Experiment. There are nine reactions you can explore on your own. Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much SO2 would you need to completely react with 6.00 g of O2 such that all reactants could be consumed
Chemistry
1 answer:
Step2247 [10]3 years ago
5 0

Answer: Thus 24.0 g of SO_2 would be needed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}      

\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles

2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)  

According to stoichiometry :

1 mole of O_2 require = 2 moles of SO_2

Thus 0.1875 moles of O_2 will require=\frac{2}{1}\times 0.1875=0.375moles  of SO_2  

Mass of SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g

Thus 24.0 g of SO_2 would be needed to completely react with 6.00 g of O_2 such that all reactants could be consumed.

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