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How do you calculate the volume of 66.5 g of carbon monoxide at STP?

Answer 1

Jeremiah Adams

#"Volume" = 53.2dm^3#

Okay, a straightforward method of comparison.

#CO rarr STP#

#"Molar mass of" color(white)xCO = 12 + 16 = 28gmol^-1#

Let the volume be represented as #x#

Consequently,

#28gmol^-1 rarr 22.4moldm^-3#

#66.5g rarr xdm^3#

multiplication by cross;

#28 xx x = 22.4 xx 66.5#

#28x = 1489.6#

#x = 1489.6/28#

#x = 53.2dm^3#

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Answer 2

Cooper Anderson

To calculate the volume of a gas at STP (Standard Temperature and Pressure), you can use the ideal gas law, which states: ( V = \frac{nRT}{P} ). At STP, ( T = 273.15 ) K and ( P = 1 ) atm. The molar mass of carbon monoxide (CO) is approximately 28.01 g/mol. So, first, you need to find the number of moles of CO by dividing the given mass by the molar mass. Then, plug the values into the ideal gas law equation to find the volume. Thus:

( n = \frac{66.5 , \text{g}}{28.01 , \text{g/mol}} )

( n \approx 2.374 , \text{mol} )

( V = \frac{2.374 , \text{mol} \times 0.0821 , \text{L} \cdot \text{atm/mol} \cdot \text{K} \times 273.15 , \text{K}}{1 , \text{atm}} )

( V \approx 62.4 , \text{L} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.