Question

What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?

Answer 1

Answer: The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

$\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol$

The given chemical equation follows:

$CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)$

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = $\frac{3}{1}\times 0.840=2.52mol$ of water

The ideal gas equation is given as:

$PV=nRT$ .......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank = $500^oC=[500+273]K=773K$

Putting values in equation 2, we get:

$1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L$

Hence, the volume of oxygen gas is 91.4 L.

Answer 2

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

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