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I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g) Lets start with the equation above. We would normally balance the equation, but in this case, we don't need to since it already is balanced. Take the coefficients in order to create a ratio: 1:5:5:1. Now, first, let's take the 80g of Iodine (V) Oxide and divide it by the element's atomic mass: 80.0/333.861 = 0.240 Now, multiply 0.2 by the ratio of Iodine to Iodine (V) Oxide. 0.240*1/1=0.2 Multiply 0.2 by molar mass of I2 0.2 * (126.9+126.9) = 60.8g ----------------------------
Now that you have the first part of your answer, now its time to get the 2nd part. First, we need to convert the 28g of CO into moles by dividing by 28. 28/28.0104 = 1 Next, let's multiply the 1 by (1/5), since that's the I2O5 to CO ratio. The answer we get by making this mole-mole conversion between the two compounds is: 1/5 or 0.2. Next, all we have to do is take the 0.2 and multiply by the molar mass of I2, which is 253.8.
253.8*0.2 = 50.8 50.8 is the answer to the 2nd part of this answer. ----------------------------- Next, we must take a look at both the 1st answer we got and the 2nd, and we must compare them side by side: 60.8g I2O5 v. 50.8g CO. Since there is less of CO, CO will run out faster than I2O5 during the reaction, and therefore, it is the limiting reagent. -------------------------
As for the % yield, that's easy. The formula for % yield is (actual yield) / (theoretical yield) * 100. But first, we must find the actual yield of I2, which can be found by taking the given # of moles and multiplying that by its molar mass. Once we have that, we just need to plug in the numbers and we'll receive our percent yield. (40.6/50.8)*100 = 79.9% And that's all that you need to know for now
Thank You so much for helping me prepare for my test. [Licking his Lips] It's such a shame that I am still going to have to eat you
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