mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
One mole of hydrogen peroxide contains 6.02 x 10^23 molecules of hydrogen peroxide. And each molecule contains 4 atoms, so the answer is 4 x 6.02 x 10^23.
<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
