Window glass is typically made by mixing soda ash (Na2CO3), limes...

Question

Window glass is typically made by mixing soda ash (Na2CO3), limestone (CaCO3), and silica sand (SiO2) and then heating to 1500 °C to drive off CO2 from the (Na2CO3) and CaCO3. The resultant glass consists of about 12% Na2O by mass, 13% CaO by mass, and 75% SiO2 by mass. How much of each reactant would you start with to prepare 0.35 kg of glass?

Accepted Answer

Welcome back everyone. We're told that fire clay is a ceramic usually made by mixing aluminum, three hydroxide, silicon dioxide and magnesium hydroxide. The resulting mixture is then heated to 1,100°C to form aluminum three oxide and magnesium oxide. By the removal of water. The fire clay product has a mass percent composition of 45% aluminum, three oxide, 50% silicon dioxide and 5% magnesium oxide, calculate the amount of each starting material required to create 200 g of fire clay. So because we know each of these re agents or compounds make up our fire clay. We want to figure out based on that mass percent or sorry, based on this mass of fire clay, given these mass percent of each of these compounds, how many grams of each of these compounds form this fire clay. So we're going to say from mass percent From the prompt. We have 45% of our aluminum three oxide, We have 50% of our silicon dioxide and we have 5% of our manganese oxide. Or sorry, magnesium oxide there. And so we're going to take the 200 g of our fire clay from the prompt and we're going to multiply by the mass percent given for each of these compounds as decimals. So we're gonna interpret 45% as 450.45. We're gonna interpret 50% of our silicon dioxide as 500.5 or 0.50 and then we're going to interpret our percent of magnesium oxide as 0.5 as a decimal. And so now getting each of our composition of these compounds and grams We're going to have an equivalent of 90 g of our aluminum three oxide. We're going to have an equivalent of 100 g of our silicon dioxide And we're going to have an equivalent of 10 g of our magnesium oxide. And so right now we've only figured out one of our first answers which is our composition of our silicon dioxide that makes up the fire clay, which according to the prompt again, is 50%. So we need to figure out out of the other 50% how much of our first aluminum three oxide and then second magnesium oxide makes up our fire clay. So we're going to have to go through some geometry. And we also need to write out equations for how our aluminum three oxide. Sorry, aluminum three oxide and our magnesium oxide are formed from magnesium hydroxide and aluminum hydroxide. So let's begin with aluminum hydroxide. So Aluminum three oxide formed from, we'll say from aluminum hydroxide. We're going to figure this out by first writing out an equation where we have aluminum hydroxide as our reactant, Which forms our aluminum three oxide and we would form water as our second product. Now to bounce things out, we're going to place a coefficient of two in front of our aluminum hydroxide, which is going to give us six molds of our oxygen and hydrogen atoms on the reactant side, Where to bounce things out, we're going to place a coefficient of three in front of our water on the product side so that we now have six moles of oxygen on the product side and six moles of hydrogen on the product side for a balanced equation. So now that we have a balanced equation, we're going to figure out the mass and sorry about that. So we need to figure out the mass of our aluminum hydroxide that is required to form Our aluminum three oxide. And so to figure that out, we're going to begin with the mass of our aluminum three oxide, which we determined above as 90 g. We're going to go from g of aluminum, three oxide, two moles of aluminum, three oxide. And let's scoot this over to make some more rooms so well scoot this all over here. So going from grams to moles, we're going to recall our molar mass of aluminum three oxide from our periodic table, which we see as 101.96 g for one mole of aluminum. Three oxide. Now canceling our units of grams of aluminum. Three oxide. We're going to go from moles of aluminum trioxide. two moles of our aluminum hydroxide where we get from the bounced equation are molar ratio between aluminum hydroxide and aluminum trioxide, which we see based on our coefficient is a 2 to 1 molar ratio. So we have two moles of aluminum hydroxide four and let's make that meter. So two moles of aluminum hydroxide for one mole of aluminum three oxide, canceling out moles of aluminum three oxide. We're now going to get our mass of aluminum hydroxide required to make aluminum trioxide by going from moles of aluminum hydroxide, two g of aluminum hydroxide. Where from our periodic table, we see that aluminum hydroxide has a molar mass equal to Sorry, so 78g of our aluminum hydroxide equivalent to one mole of aluminum hydroxide. So canceling out moles of aluminum hydroxide were left with grams of aluminum hydroxide. And we know that from this calculation, our mass of aluminum hydroxide required to form our aluminum 33 oxide. We're going to have a mass of 138 g. That is required To get that 90 g of our aluminum three oxide. And now, following the same steps for our magnesium oxide, we have magnesium oxide formed from our magnesium hydroxide by the following equation where we have magnesium hydroxide which loses water to form our product, magnesium oxide and water as a product. And as our coefficients are as they are, we just we don't need to add any new ones. Everything is balanced here. So we can confirm that this equation is balanced. We checked the atoms on both sides of our equation. So now figuring out the mass of our magnesium hydroxide that is required to form our 10 g of magnesium oxide. We're going to again do this stock geometry step and let's make sure we have enough room. So we'll scoot this over here, Beginning with the 10 g of our magnesium oxide. Getting rid of grams of magnesium oxide to go two moles of magnesium oxide we recall from the periodic table are molar mass of magnesium oxide, which is 40.30 g for one mole, canceling out our grams of magnesium oxide and going from moles of magnesium oxide, two moles of our magnesium hydroxide. And sorry, this is 0H two. Here We see that according to our balanced reaction, we have a 1-1 molar ratio. So now canceling out moles of magnesium oxide and going from moles of magnesium hydroxide. We'll carry this down below. We want to get rid of moles of magnesium hydroxide And get two g of magnesium hydroxide utilizing our molar mass from the periodic table where we see we have 58.32 g of magnesium hydroxide equivalent to one mole of magnesium hydroxide, allowing us to now cancel out moles of magnesium hydroxide in leaving us with our final unit as grams. This is going to yield a mass of magnesium hydroxide required to make our 10 g of magnesium oxide being 15 g of magnesium hydroxide. And so for our last two final answers, we have determined that from our starting materials, we need 138 g of the aluminum hydroxide starting material to form our 90 g of aluminum, three oxide. And our third and final answer is that we need 15 g of our starting material magnesium hydroxide to form our 10 g of magnesium oxide to ultimately form enough of our fire clay, so it's highlighted in yellow are our three final answers. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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