1.) For the reaction, calculate how many moles of the product form when 0.032mol
ID: 854368 • Letter: 1
Question
1.) For the reaction, calculate how many moles of the product form when 0.032mol of K completely reacts. Assume that there is more than enough of the other reactant. 4K(s)+O2(g)?2K2O(s). Express your answer using two significant figures.
2.) Hydrochloric acid can dissolve solid iron according to the following reaction.
Fe(s)+2HCl(aq)?FeCl2(aq)+H2(g)
A.) What minimum mass of HCl in grams would you need to dissolve a 2.6g iron bar on a padlock? Express your answer using two significant figures.
B.) How much H2 would be produced by the complete reaction of the iron bar? Express your answer using two significant figures.
3.) Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:
2Mg(s)+O2(g)?2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 11.7g of MgO are collected.
A.) Determine the limiting reactant for the reaction. Express your answer as a chemical formula.
Explanation / Answer
1.)
4K(s)+O2(g) --------> 2K2O(s)
From the reaction it is clear that 4 mol of K gives 2 moles of K2O.
So, 0.032 mol if K gives 2 x 0.032 mol = 0.064 mol of K2O.
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2.)
Fe(s)+2HCl(aq) ---------------> FeCl2(aq)+H2(g)
A.)
Mass of iron bar = 2.6 g.
Moles of iron bar = 2.6g / 55.84 = 0.046 mol.
From the chemical reaction, it is clear that, 1 mol of Fe reacts with 2 mol of HCl.
So, moles of HCl required = 2 x 0.046 mol
= 0.092 mol of HCl
Mass of HCl = 0.092 mol x 36.5 g/mol = 3.358 g
B.)
Mass of iron bar = 2.6 g.
Moles of iron bar = 2.6g / 55.84 = 0.046 mol.
From the chemical reaction, it is clear that, 1 mol of Fe reacts with 1 mol of H2.
So, 0.046 mol of Fe gives 0.046 mol of H2.
Mass of H2 = 0.046 mol x 2 g/mol = 0.092 g of H2.
3.)
2Mg(s)+O2(g) --------------> 2MgO(s)
Given, 10.1 g Mg.
Moles of Mg = 10.1 g / 24.3 = 0.42 mol.
Given, 10.5 g of O2.
Moles of O2 = 10.5 g / 32 g/mol = 0.33 mol.
Finding the limiting reagent:
We need to divide the moles obtained with the respective stoichiometric coefficients.
Mg = 0.42 mol / 2 = 0.21 mol
O2 = 0.33 mol / 1 = 0.33 mol.
As, Mg has less number of moles, Mg is the limitig reagent.
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