Calculate the grams of so2 gas present at stp in a 5.9 l container.

Chemistry

1 answer:

nata0808 [166]3 years ago

7 0

Answer: The mass of sulfur dioxide gas at STP for given amount is 16.8 g

Explanation:

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas.

So, 5.9 L of volume will be occupied by = $\frac{1mol}{22.4L}\times 5.9L=0.263mol$

Now, to calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sulfur dioxide gas = 0.263 mol

Molar mass of sulfur dioxide gas = 64 g/mol

Putting values in above equation, we get:

$0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g$

Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g

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