Dry ice (solid CO2) has occasionally been used as an “explosive” ...

Question

Dry ice (solid CO2) has occasionally been used as an “explosive” in mining. A hole is drilled, dry ice and a small amount of gunpowder are placed in the hole, a fuse is added, and the hole is plugged. When lit, the exploding gunpowder rapidly vaporizes the dry ice, building up an immense pressure. Assume that 500.0 g of dry ice is placed in a cavity with a volume of 0.800 L and the ignited gunpowder heats the CO2 to 700 K. What is the final pressure inside the hole?

Accepted Answer

Welcome back everyone. We're told that mining has occasionally used zinc carbonate as an explosive drilling a hole, inserting zinc carbonate, a trace amount of gunpowder a fuse and plugging the hole are the steps when ignited. This creates a tremendous amount of pressure. Assume that 780g of zinc carbonate Are put in a cavity that has a volume of .350 L and that the zinc carbonate is heated to 650 Kelvin by the gunpowder that has been ignited. How much pressure is inside the hole at the end. Our first step is to write out the reaction for what's happening. So we have our zinc carbonate solid, which when heated, is going to decompose to zinc oxide as well as carbon dioxide gas. Now based on this equation, we would see that our only gas involved is carbon dioxide gas, which is our product here and to determine the pressure inside the hole. At the end, we want to figure out based on our moles of gas left, what are pressure would be. And so we're going to recall our formula where pressure times volume is equal to our most of our gas times R gas constant R times our temperature in kelvin set equal to pressure times volume. And so, using this formula, we first need to begin by finding out how much moles of C. 02 we have. So we're going to begin by making note of our molar mass of our zinc carbonate from our periodic table And we would see that for zinc carbonate, we have a molar mass of 125.4 g per mole from the periodic table. So using this molar mass, we're going to first get moles of zinc carbonate by taking our Mass of zinc carbonate given from the prompt being 780 g Of zinc carbonate. And multiplying it by our molar mass where we are interpreting it as 125.4 g of zinc carbonate equivalent to one mole of zinc carbonate. So now canceling out grams of zinc carbonate. We have multi zinc carbonate as our final unit. And this is going to yield our molten zinc carbonate of 6.2201 moles. So now that we have moles of our reactant, we're going to assume that all of our zinc carbonate reactant decomposes. And so therefore with this assumption, we can say that our moles of our reactant zinc carbonate should be equivalent to our moles formed of our product carbon dioxide gas. And so We can say then then our moles of carbon dioxide should also equal 6.2201 moles. And so now with this understanding we can go into our pressure equation, we're isolating for pressure, we're going to say that pressure is equal to the mold of our gas times r gas constant R times temperature divided by our volume. We're plugging in what we know, we have our moles of our carbon dioxide, which is our only gas from our equation here being and will circle that in pink actually. So plugging that in, we have that value as 6.2201 most. So this is for our term end for moles of our gas moles of CO2. This is multiplied by r. Gas constant R Which we want to recall is 0.08206. And sorry, let's actually carry this down so that we have enough room for everything. So this is going to be multiplied again by our gas constant. R 0.08206 with units of leaders times most or sorry, Leaders times A T. M's divided by moles times kelvin And then multiplying this by our temperature given in the prompt as 650 Kelvin. We then want to divide by our volume which was given in the prompt also as 0.350 liters. So now canceling out our units, Let's get rid of kelvin with kelvin and numerator. Leaders with leaders in the denominator and moles. And we're left with a T. M. As our final unit of pressure. And this is going to yield a result for our pressure equal to about 948 ATMs of pressure. And this would complete this example as our final answer for our pressure inside the hole at the end of the reaction. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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