2 K(s) + Cl 2 (g) → 2 KCl(s)

Hello. Hello everyone. Today we have a phone problem for this reaction. Calculate the mass in grams with a product that forms when 15.39 g of the underlined reactant completely reacts assume that there is more than enough of the other reactant. So we are being asked to calculate the mass of the gram of the product that forms when 15.39 g of the strontium forms. So we always start with our given which was 15.39 g of our strontium. And then we will multiply by the molar mass. So for our units to cancel, we need to put grams on the denominator. And our new are the moles on the numerator. So we say that one mole of our strontium is equal to 87.62 g of strontium. And this is according to the periodic table. And then the next thing that we have to multiply to get to Graham's again is to cancel our moles. So you have to multiply by the multiple ratio. Can I do that? We have to put the moles of rat, the moles of strontium on the denominator. So we have two moles of strontium given the underlined moles in our question. And then we see how many moles that we need to get to. So the product. So strontium oxide, we see that we have two moles of that. So we say we have two moles of strontium oxide in our numerator. And then lastly to get from moles to grams, we multiply strontium oxide's molar mass. And we see that one mole of strontium oxide and the denominator is equal to the molar mass of strontium oxide. When we take the molar mass or the mass in the table of strontium and oxygen, and we add them up to get 103.62 g. And so if we cancel up all of our units, we will arrive at a mass of 18.20 g of strontium oxide or anti choice B. And with that, we have solved the problem overall, I hope this helped. And until next time.

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2 K(s) + Cl₂(g) → 2 KCl(s)

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2 K(s) + Cl2(g) → 2 KCl(s)

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2 K(s) + Cl₂(g) → 2 KCl(s)