Question

The liquid hydrogen/oxygen rocket will burn for about 8 minutes at a rate of 90,000 gallons of liquid hydrogen per minute. The density of liquid hydrogen is 70.85 g/L. How many moles of water vapor will be generated during the 8 minute burn of this rocket?

Answer 1

The number of moles of the water that was evaporated is  12083467.5 moles of water.

What moles of water is evaporated?

We know that the number of moles that we have can be gotten as the ratio of the mass to the number of moles of the object. Now we know that from the question, we have the volume of the hydrogen to be 90,000 gallons. We have to convert this volume to liters.

Given that;

1 gallon = 3.79 L

90,000 gallons  = 90,000 gallons  * 3.79 L/1 gallon

= 341100 L

Then we have the density of the hydrogen to be 70.85 g/L

Mass of the hydrogen =  70.85 g/L *  341100 L

= 24166935 g

Number of moles of the hydrogen reacted = 24166935 g/2 g/mol

= 12083467.5 moles

If 2 mole of hydrogen gives 2 moles of water

12083467.5 moles of hydrogen gives 12083467.5 moles of water

Learn more about formation of water:brainly.com/question/9524396

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