Question 10 1 pts
Given the following balanced chemical equation:
2C2H10 + 902 +4CO2 + 10H2O
How many grams of C2H10 are needed to produce 150. g of water?
396 g
15.8 g
O 1420
O 56.8 g

56.8 g

Explanation:

Your first conversion factor is the molar mass of H2O to convert from grams of water to moles. Your second conversion factor is the mole ratio of 2 mol C2H10 for every 10 mol H2O. Your last conversion factor is the molar mass of C2H10 to convert from moles back to grams.

150. g H2O • (1 mol H2O / 18.02 g H2O) • (2 mol C2H10 / 10 mol H2O) • (34.1 g C2H10 / 1 mol C2H10) = 56.8 g C2H10

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o