Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.22 g of ethane is mixed with 15. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Round your answer to significant digits.

The remaining mass of ethane is 3.19 grams

Explanation:

Step 1: Data given

Mass of ethane = 7.22 grams

Mass of oxygen = 15.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = 7.22 grams / 30.07 g/mol

Moles ethane = 0.240 moles

Step 4: Calculate moles oxygen

Moles oxygen = 15.0 grams / 32.0 g/mol

Moles oxygen = 0.46875 moles

Step 5: Calculate limiting reactant

For 2 moles ethane we need 7 moles oxygen to produce 4 moles CO2 and 6 moles H2O

Oxygen is the limiting reactant. It will completely be consumed (0.46875 moles). Ethane is in excess. There will react 0.46875 / 3.5 = 0.134 moles

There will remain 0.240 - 0.134 = 0.106 moles

Step 6: Calculate mass of remaining ethane

Mass ethane = 0.106 moles * 30.06 g/mol

Mass ethane = 3.19 grams

The remaining mass of ethane is 3.19 grams

answer; protein;

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.