In an experiment, 8.50 g of methane, CH4, was reacted with 15.9 g of oxygen gas to produce carbon dioxide and water. Determine the percentage yield if 9.77 g of carbon dioxide was obtained in the lab.

89.3 %

Explanation:

M(CH4) = 12+ 4*1 = 16 g/mol

M(O2) = 2*16 = 32 g/mol

M(CO2) = 12 + 2*16 = 44 g/mol

8.50 g * 1 mol/16 g = 0.5313 mol CH4

15.9 g * 1 mol/32 g = 0.4969 mol O2

9.77 g * 1 mol/44 g = 0.2220 mol CO2

1)                                  CH4        + 2O2 > CO2 + 2H2O

from reaction             1 mol          2 mol

given                       0.5313 mol   (0.4969 mol)

1 mol CH4              --- 2 mol O2

0.5313 mol  CH4  --- x mol O2

x= 2*0.5313 = 1.0626 mol O2

We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.

2)                               CH4        + 2O2 > CO2 + 2H2O

from reaction                             2 mol        1 mol

given                                      0.4969 mol   x mol

x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield

3)

Practical yield CO2 = 0.2220 mol

Theoretical yield CO2 = 0.2485 mol

% yield = (0.2220/0.2485)*100% = 89.3 %

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: