7. Tungsten (W) metal, which is used to make incandescent bultb flaments, s prod

Question

7. Tungsten (W) metal, which is used to make incandescent bultb flaments, s produced by the reaction WO, + 3H, 3H0 + w How many grams of H2 are needed to produce 4.60 g of W 18. A mixture of hydrazine (N2H4) and hydrogen peroxide (H202) is used as a fu for rocket engines. These two substances react as shown by the equation What mass of N,H, in grams, is needed to react completely with 38.27 g of H,0 19. The theoretical yield of product for a particular reaction is 57.23 g. A very careless student obtained only 20.12 g of product after carrying out this reaction. What is the percent yield that this student obtained? 20. In an experiment designed to produce calcium oxide by the chemical reaction 2Ca + O2 2CaO 62.39 g of CaO was obtained out of a possible 70.1 g of Cao (a) What is the theoretical yield of CaO? (b) What is the actual yield of CaO? (c) What is the percent yield of Ca0?

Explanation / Answer

Solution:- All these are stoichiometry problems and so these are solved by using the mole ratio from balanced equation.

(17) Given balanced equation is, WO3 + 3H2 -------> 3H2O + W

we have been given with grams of W and asked to calculate the grams of H2 required.

First we will convert the grams of W into moles and for this we divide the grams by molar mass. Then we multiply these moles by mole ratio which is 3moles of H2/1 mole of W to get the moles of H2 and finally these moles are converted to grams of H2 by multipling the molar mass. So, all theese steps together would be as...

4.60 g of W x (1 mole of W/183.84 g of W) x (3moles of H2/1 mole of W) x (2.02 g of H2/1mole of H2 ) = 0.152 g of H2.

(18) Given balanced equation is..   N2H4 + 2H2O2 -------> N2 + 4H2O

we are asked to calculate the grams of N2H4 for the given grams of H2O2. This is same as problem #17. So, let's make the set up as..

38.27 g of H2O2 x ( 1 mole of H2O2/ 34.02 g of H2O2) x (1 mole of N2H4/ 2 mole of H2O2) x (32.06 g of N2H4/1 mole of N2H4) = 18.03 g of N2H4. (answer)

(19) percent yield = (actual yield/theoretical yield) x 100

Theoretical yield is given as 57.23 g and the actual yield is given as 20.12 g. So, let's plug in these values in the formula.

percent yield = (20.12/57.23) x 100 = 35.16% (answer)

(20) Given balanced equation is..   2Ca + O2 --------> 2CaO

62.39 g of Cao was obtained out of a possible 70.1 g of CaO. It means 70.1 g is the theoretical yield and 62.39 g is the actual yield since actual yield is always less than theoretical yield because theoretically we assume 100% reaction where in actual no reaction is 100%. So,

(a) Theoretical yield is 70.1 g.

(b) Actual yield is 62.39g.

(c) percent yield = (62.39/70.1) x 100 = 89.0%

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

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