when 50.0 grams of silicon dioxide is heated with an excess of carbon, 32.2 gram
ID: 690964 • Letter: W
Question
when 50.0 grams of silicon dioxide is heated with an excess of carbon, 32.2 grams carbonide along with carbon monoxide. what is the percent yield of silicon dioxide? how many grams of carbon monoxide gas are produced?Explanation / Answer
The equation for the reaction is: . SiO2 + 3 C-----> 2 CO + SiC . 1 mole of silicon dioxide gives two moles of CO and one mole ofSiC. Moles of SiO2 = mass / molar mass = 50.0 g / 60.08g/mol = 0.832 moles . Moles of product SiC = 32.2 g / 40.1 g/mol = 0.803 moles . Since 1 mole of SiO2 gives 1 mole of Si C, the theoretical yield ofSiC = 0.832 moles . I suppose the question is about the percent yieldof silicon carbide and not silicon dioxide . Percent yield of silicon carbide = ( actual yield / theoreticalyield) * 100 = ( 0.803 / 0.832) * 100 = 96.5 % . Theoretical moles of CO produced = 2* 0.832 = 1.664 moles . Actual yield = percent yield / 100 * theoretical yield = 96.5 % / 100 * 1.664 moles = 1.606 moles . mass of CO produced = moles * molar mass = 1.606 moles * 28.01 g/mol = 44.98 g
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.