1.- Silicon tetrachloride (SiCl 4 ) can be preparedby heating Si in chlorine gas

Question

1.- Silicon tetrachloride (SiCl4) can be preparedby heating Si in chlorine gas: Si(s) + 2Cl2(g) ----->SiCl4(l) In one reaction, 0.507 mole ofSiCl4 is produced. How many moles of molecular chlorinewere used in the reaction? 2.-Consider the combustion of butane(C4H10):    2C4H10(g) + 13O2(g) ---->8CO2(g) + 10H2O(l) In a particularreaction, 5.0 moles of C4H10 react with anexcess of O2. Calculate the number of moles ofCO2 formed. 3.- When baking soda (NaHCO3) is heated , itreleases carbon dioxide gas, which is responsible for the rising ofcookies, bread, etc... a) Write a balanced eqn.for thedescomposition of the compound (one of the products is Na2CO3) . b) Calculate the mass ofNaHCO3   required to produce 20.5 g ofCO2 1.- Silicon tetrachloride (SiCl4) can be preparedby heating Si in chlorine gas: Si(s) + 2Cl2(g) ----->SiCl4(l) In one reaction, 0.507 mole ofSiCl4 is produced. How many moles of molecular chlorinewere used in the reaction? 2.-Consider the combustion of butane(C4H10):    2C4H10(g) + 13O2(g) ---->8CO2(g) + 10H2O(l) In a particularreaction, 5.0 moles of C4H10 react with anexcess of O2. Calculate the number of moles ofCO2 formed. 3.- When baking soda (NaHCO3) is heated , itreleases carbon dioxide gas, which is responsible for the rising ofcookies, bread, etc... a) Write a balanced eqn.for thedescomposition of the compound (one of the products is Na2CO3) . b) Calculate the mass ofNaHCO3   required to produce 20.5 g ofCO2

Explanation / Answer

(1)              Si(s) + 2Cl2(g) -----> SiCl4(l)      1 mole of SiCl 4 is produced from 2moles of Cl 2     0.507 moles of SiCl 4 is produced from xmoles of Cl 2     x = ( 2 * 0.507 ) / 1        = 1.014 moles (2) 2C4H10(g) + 13O2(g)----> 8CO2(g) + 10H2O(l) 2 moles of C4H10 produces 8 molesof CO 2 5moles of C4H10 produces y molesof CO 2 y = ( 8 * 5 ) / 2          = 20moles (3) ( a ) 2NaHCO3 ---> Na2CO3+ CO2 + H2O ( b ) mol.wt. of NaHCO3 is = 23 + 1 + 12 + 3(16 ) = 84 g mol. wt. of CO 2 is = 12 + 2 ( 16 ) = 44 g 2 ( 84 ) g of NaHCO3 produces 44 g ofCO 2 z g of NaHCO3 produced from 20.5 g ofCO 2 z = ( 168 * 20.5 ) / 44         = 78.2727 g
In a particular reaction, 5.0 moles ofC4H10 react with an excess of O2.Calculate the number of moles of CO2 formed. y = ( 8 * 5 ) / 2          = 20moles (3) ( a ) 2NaHCO3 ---> Na2CO3+ CO2 + H2O ( b ) mol.wt. of NaHCO3 is = 23 + 1 + 12 + 3(16 ) = 84 g mol. wt. of CO 2 is = 12 + 2 ( 16 ) = 44 g 2 ( 84 ) g of NaHCO3 produces 44 g ofCO 2 z g of NaHCO3 produced from 20.5 g ofCO 2 z = ( 168 * 20.5 ) / 44         = 78.2727 g
In a particular reaction, 5.0 moles ofC4H10 react with an excess of O2.Calculate the number of moles of CO2 formed.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.