Ethane gas, or dicarbon hexahydride, undergoes combustion to produce carbon dioxide gas and water vapor.
A.) What mass of the un named reactant is required to react with moles of ethane?
B.) How many molecules of the porduct, with the smaller percent composition of the non-oxide component, are produced if the reaction only goes 57.8 percent to completion?
You didn't provide the moles of ethane. The un-named reactant is oxygen.
For the second part, here is an example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To solve these problems, we'll need to use stoichiometry and the concept of percent composition. Let's break down each question step-by-step:
A.) What mass of the unnamed reactant is required to react with moles of ethane?
To answer this question, we need to know the balanced chemical equation for the combustion of ethane:
C2H6 + O2 → CO2 + H2O
From this equation, we see that the reactant ethane (C2H6) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
First, we need to determine the moles of ethane given. If you have the moles of ethane, simply substitute that value in the following steps.
1) To convert moles of ethane to moles of the unnamed reactant, we use the stoichiometric molar ratio. From the balanced equation, we see that the coefficient of ethane is 1, and the coefficient of the unnamed reactant (let's call it X) is also 1. Therefore, the ratio is 1:1 for ethane to X.
2) Now, we need to convert the moles of X to mass. To do this, we need the molar mass of X. If you have the molar mass of X, substitute that value in the following steps.
Multiply the number of moles of X by its molar mass to get the mass of X required.
B.) How many molecules of the product, with the smaller percent composition of the non-oxide component, are produced if the reaction only goes 57.8 percent to completion?
In this question, we need to calculate the number of molecules of the product with the smaller percent composition of the non-oxide component. To do this, we can follow these steps:
1) Determine the balanced chemical equation for the combustion of ethane:
C2H6 + O2 → CO2 + H2O
From the equation, we see that for every 1 mol of ethane burned, 1 mol of carbon dioxide is produced. Therefore, the stoichiometric ratio is 1:1.
2) Calculate the theoretical yield of carbon dioxide by multiplying the moles of ethane reactant by the stoichiometric ratio.
3) Multiply the theoretical yield of carbon dioxide by the percentage of completion (57.8%) to obtain the actual yield of carbon dioxide.
4) Convert the actual yield of carbon dioxide from moles to molecules. One mole of any substance contains Avogadro's number of molecules (6.022 x 10^23).
Note: The question mentions the "smaller percent composition of the non-oxide component" without providing specific percentages. You might need additional information or specific percentages to accurately answer this part.
Remember, stoichiometry involves balancing equations and using mole ratios to convert between different units and substances. Pay close attention to coefficients in the balanced equation and ensure you have the correct units for your final answer.