Asked by Louis
How many grams of gaseous required?
If 0.260 mol of solid C and 20.0 g of solid TiO2 are reacted stoichiometrically according to the balanced equation, how many grams of gaseous Cl2 are required?
3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(l) + 2CO2(g) + 2CO(g)
Answers
Answered by
DrBob222
1. Make sure the equation is balanced.
2a. Convert 0.260 g C to moles. moles = grams/molar mass.
2b Convert 20.0 g TiO2 to moles using the same procedure.
3a. Using the coefficients in the balanced equation, convert moles C in 2a to moles Cl2.
3b. Using the same procedure, convert moles TiO2 in 2b to moles Cl2.
3c. More than likely the answer for 2a and 2b will be different. Obviously both can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value. (The material that produces the smaller value is the limiting reagent.)
4. Convert the value from 3c to grams. grams = moles Cl2 x molar mass Cl2.
2a. Convert 0.260 g C to moles. moles = grams/molar mass.
2b Convert 20.0 g TiO2 to moles using the same procedure.
3a. Using the coefficients in the balanced equation, convert moles C in 2a to moles Cl2.
3b. Using the same procedure, convert moles TiO2 in 2b to moles Cl2.
3c. More than likely the answer for 2a and 2b will be different. Obviously both can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value. (The material that produces the smaller value is the limiting reagent.)
4. Convert the value from 3c to grams. grams = moles Cl2 x molar mass Cl2.
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