Answer:
194 g/mol.
Explanation:
Hello,
In this case, one first must compute the mass of each element as shown below:
Next, the corresponding moles:
Then, each element's subscripts is found to be:
Therefore, the empirical formula is:
Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:
Which has a molar mass of 194 g/mol being correctly contained in the given interval.
Best regards.
Ethene react with oxygen at a molar ratio:
Convert the quantity of each reactant supplied to number of moles of particles:
The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, and
(as seen in the balanced chemical equation) shall satisfy the relationship
In other words,
Evaluating the expression with data given in the question yields approximately
, which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.
The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio
,