Calculate the mass of water produced when 1.36 g of butane reacts with excess oxygen.

Mass of water produced= 1.8 g

Explanation:

Given data:

Mass of water produced = ?

Mass of butane = 1.36 g

Mass of oxygen = excess

Solution:

Chemical equation:

2C₄H₁₀ +13 O₂       →   8CO₂ + 10H₂O

Number of moles of butane:

Number of moles = mass/molar mass

Number of moles = 1.36 g/ 58.12 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of butane with water.

                 C₄H₁₀         :        H₂O

                  2              :           10

                 0.02          :         10/2×0.02 = 0.1 mol

Mass of water produced:

Mass = molar mass × molar mass

Mass = 0.1 mol × 18 g/mol

Mass = 1.8 g

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.