If starting with 18.2 grams of carbon, how many moles oftetraboron tricarbide, B

Question

If starting with 18.2 grams of carbon, how many moles oftetraboron tricarbide, B4C3, can be theoretically formed? 2 B2O3 (s) + 6 C (s) ----->  B4C3 (s)  + 3 CO2 (g) Thanks for your help! If starting with 18.2 grams of carbon, how many moles oftetraboron tricarbide, B4C3, can be theoretically formed? 2 B2O3 (s) + 6 C (s) ----->  B4C3 (s)  + 3 CO2 (g) Thanks for your help!

Explanation / Answer

The stoichiometric equation is 2 B2O3 (s) + 6 C (s) ----->   B4C3(s)  + 3 CO2 (g)                    6X12=72g           4(10.81)+36                                                    =79.24g                     18.2g                     Xg                     X=(79.24 X 18.2 ) / 72 = 20.03011g here as per the equation 72 grams of carbon producing79.24 grams of Boron carbide     we calculated for 18.2g carbon by using aboveratio.

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