Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.82 g of ethane is mixed with 9.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits

4.79 g of water

Explanation:

From the reaction equation;

C2H6(g) + 7/2O2(g) > 2CO2(g) + 3H2O(g)

Next we convert the given masses of reactants to moles of reactants.

For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles

For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles

Next we determine the limiting reactant, the limiting reactant yields the least amount of product.

For ethane;

From the reaction equation,

1 mole of ethane yields 3 moles of water

0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water

For oxygen;

3.5 moles of oxygen yields 3 moles of water

0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water

Hence oxygen is the limiting reactant.

Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.