Question

what is the volume of 8.8g of carbon dioxide at STP?​

Answer 1

Answer:

Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K–1 mol–1. Hence, the volume occupied is 5.05 L.

Explanation:

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Answer 2

Answer:

The volume of the 8.8 g carbon dioxide gas at STP is 4.48 liters.

Explanation:

Here we have been given that the mass of the carbon dioxide is 8.8 g. We have to find the volume of the carbon dioxide gas here. Now as we know the number of moles of the carbon dioxide gas is equal to the ratio of the given mass of the gas to the molecular or molar mass of the gas. Hence we calculate its number of moles as below:

The molar mass of the oxygen (O) atom = 16 g

Molar mass of carbon (C)atom = 12 g

∴ The molar mass of Carbon dioxide $CO_{2}$

= 12g + 2 × 16 g

= 12 g + 32g

= 44g

Therefore,

no. of moles of carbon dioxide(n)  = given mass of carbon dioxide(m) / molar mass of carbon dioxide(M)

m = 8.8 g M = 44 g

∴ n = $\frac{8.8g}{44g}$

⇒ n = 0.2 moles

Now we know that at STP one mole of the gas has a volume which is equal to 22.4 liters/mol

Therefore we have,

The volume of 1 mole of $CO_{2}$ = 22.4 liters

∴ The volume of 0.2 moles of $CO_{2}$ = 0.2 × 22.4 litres

= 4.48 liters

Hence the volume of 8.8 g of carbon dioxide is found to be 4.48 liters.