The combustion of methyl alcohol in an abundant excess of oxygen follows the equation:

2CH3OH + 3O2? 2CO2 + 4H2O

When 3.85 g of CH3OH was mixed with 6.15 g of O2 and ignited, 3.65 g of CO2 was obtained. What was the percentage yield of CO2?

The % yield of CO2 is 69.1 %

Explanation:

Step 1: Data given

Mass of CH3OH = 3.85 grams

Mass of O2 = 6.15 grams

Mass of CO2 obtained = 3.65 grams

Molar mass of CH3OH = 32.04 g/mol

Molar mass O2 = 32.00 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2CH3OH + 3O2→ 2CO2 + 4H2O

Step 3: Calculate moles  CH3OH

Moles CH3OH = mass CH3OH / molar mass CH3OH

Moles CH3OH = 3.85 grams / 32.04 g/mol

Moles CH3OH = 0.120 moles

Step 4: Calculate moles O2

Moles O2 = 6.15 grams / 32.00 g/mol

Moles O2 = 0.192 moles

Step 5: Calculate limtiting reactant

CH3OH is the limiting reactant. It will completely be consumed ( 0.120 moles).  O2 is in excess. There will react 3/2 * 0.120 = 0.180 moles

There will remain 0.192 - 0.180 = 0.012 moles

Step 6: Calculate moles CO2

For 2 moles CH3OH we need 3 moles O2 to produce 2 moles CO2 and 4 moles H2O

For 0.120 moles CH3OH we'll have 0.120 moles CO2

Step 7: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.120 moles * 44.01 g/mol

Mass CO2 = 5.28 grams

Step 8: Calculate % yield

% yield = (actual mass / theoretical mass) *100%

%yield = ( 3.65 grams / 5.28 grams ) *100%

% yield = 69.1 %

The % yield of CO2 is 69.1 %

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.