Use the following equation to determine how many grams of o2 are required to react with 22.50g of c7h16?

c7h16 + 11o2 --> 7co2 + 8h2o

79.2 grams of oxygen gas are required to react with 22.50 grams of n-heptane.

Explanation:

Mass of n-heptane = 22.50 g

Moles of n-heptane =

According to reaction 1 mole of n-heptane reacts with 11 moles of oxygen gas. then 0.2250 moles of n-heptane will react with:

of oxygen gas

Mass of 2.475 moles of oxygen gas= 2.475 mol × 32 g/mol= 79.2 g

79.2 grams of oxygen gas are required to react with 22.50 grams of n-heptane.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).