B, they then interpret that data to find their answers
<span>1 mole glucose gives 2 moles of ethanol
moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles
so moles of ethanol produced = 2* 13.32 = 26.64 moles
weight of ethanol 26.64 * 46.07
=1227.30 gm or 1.23 Kg</span>
Answer:
43.72
Explanation:
that is the answer hope u liked it and I did this already along time ago
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:
![m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2](https://tex.z-dn.net/?f=m_%7BCO_2%7D%5E%7Btheoretical%7D%3D10gCO%2A%5Cfrac%7B1molCO%7D%7B28gCO%7D%2A%5Cfrac%7B2molCO_2%7D%7B2molCO%7D%20%20%2A%5Cfrac%7B44gCO_2%7D%7B1molCO_2%7D%5C%5C%5C%5C%20m_%7BCO_2%7D%5E%7Btheoretical%7D%3D16gCO_2)
Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:
![Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B10g%7D%7B16g%7D%20%2A100%5C%25%5C%5C%5C%5CY%3D62.5%5C%25)
Best regards!