1)The reaction between solid sodium and iron(III) oxide, Fe2O3(s), is one in a series of reactions that occurs when an automobile air bag inflates.
6Na(s) + Fe2O3(s) -> 3Na2o(s) + 2Fe(s)
If 100.0 g of iron(III) oxide are used in this reaction, what mass of solid iron will be produced?
2)Sliver nitrate , AgNO3(aq), reacts with calcium chloride, CaCl2(aq), in the following reaction:
2AgNO3(aq) + CaCl2(aq) -> 2AgCl(s) + Ca(NO3)2(aq)
There are 7.000 mol of each reactant present.
a.What is the mass of the excess reactant?
b.What is the mass of the limiting reactant?
c.What are the masses of each product that forms?
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1) To determine the mass of solid iron produced, we need to use the given molar mass of Fe2O3 and the stoichiometry of the balanced equation.
a) Calculate the molar mass of Fe2O3:
Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
b) Calculate the number of moles of Fe2O3 used:
Number of moles = mass / molar mass
Number of moles = 100.0 g / 159.69 g/mol = 0.6262 mol
c) Using the stoichiometry of the balanced equation:
From the balanced equation, we can see that the mole ratio of Fe2O3 to Fe is 1:2.
Therefore, the number of moles of Fe produced will be twice the number of moles of Fe2O3 used:
Number of moles of Fe = 2 * 0.6262 mol = 1.2524 mol
d) Calculate the mass of Fe produced:
Mass = number of moles * molar mass
Mass = 1.2524 mol * 55.85 g/mol = 69.89 g
Therefore, the mass of solid iron produced is 69.89 g.
2) To determine the mass of the excess reactant and the limiting reactant, we need to compare the molar ratios of the reactants to the balanced equation.
a) Calculating the molar mass of AgNO3 and CaCl2:
AgNO3 = 107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol) = 169.87 g/mol
CaCl2 = 40.08 g/mol + 2(35.45 g/mol) = 110.98 g/mol
b) Calculate the number of moles of AgNO3 and CaCl2:
Number of moles = number of mol x molarity (assuming the given amount is molarity)
Number of moles of AgNO3 = 7.000 mol
Number of moles of CaCl2 = 7.000 mol
c) Using the stoichiometry of the balanced equation:
From the balanced equation, we can see that the mole ratio of AgNO3 to AgCl is 2:2, and the mole ratio of CaCl2 to AgCl is 1:2.
Therefore, the mole ratio of AgNO3 to CaCl2 is 2:1.
Since the mole ratio is 2:1, we have twice the number of moles of AgNO3 compared to CaCl2.
Number of moles of AgNO3 in excess = 7.000 mol - (7.000 mol / 2) = 3.500 mol
d) Calculate the mass of the excess reactant:
Mass = number of moles * molar mass
Mass of AgNO3 in excess = 3.500 mol * 169.87 g/mol = 594.55 g (rounded to two decimal places)
e) Determine the limiting reactant by comparing the number of moles of AgNO3 and CaCl2 used. Since the mole ratio is 2:1, we have twice the number of moles for AgNO3.
Number of moles of CaCl2 used = 7.000 mol / 2 = 3.500 mol
f) Calculate the mass of the limiting reactant:
Mass of CaCl2 used = 3.500 mol * 110.98 g/mol = 388.43 g (rounded to two decimal places)
g) Now, calculating the masses of the products formed:
From the balanced equation, we can see that the mole ratio of AgCl to Ca(NO3)2 is 2:1.
Therefore, the number of moles of AgCl and Ca(NO3)2 formed will be the same:
Number of moles of AgCl = 3.500 mol
Number of moles of Ca(NO3)2 = 3.500 mol
Calculate the mass of AgCl:
Mass = number of moles * molar mass
Mass of AgCl = 3.500 mol * (107.87 g/mol + 35.45 g/mol) = 543.55 g (rounded to two decimal places)
Calculate the mass of Ca(NO3)2:
Mass = number of moles * molar mass
Mass of Ca(NO3)2 = 3.500 mol * (40.08 g/mol + 2(16.00 g/mol)) = 246.84 g (rounded to two decimal places)
a) The mass of the excess reactant, AgNO3, is 594.55 g.
b) The mass of the limiting reactant, CaCl2, is 388.43 g.
c) The mass of AgCl formed is 543.55 g, and the mass of Ca(NO3)2 formed is 246.84 g.
To solve these problems, we need to use stoichiometry, which involves converting between the masses of reactants and products using the balanced chemical equation. The steps involved are as follows:
1) Mass-Mole Conversion:
a) Determine the molar mass of Fe2O3:
Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe2O3 = 2(Fe) + 3(O) = 2(55.845 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
b) Convert the mass of Fe2O3 to moles:
Moles of Fe2O3 = mass (g) / molar mass (g/mol)
Moles of Fe2O3 = 100.0 g / 159.69 g/mol
2) Mole-Mole Conversion:
Using the balanced chemical equation, we can determine the mole ratio between Fe2O3 and Fe:
6 moles of Na: 2 moles of Fe2O3
Since the coefficient of Fe2O3 is 1, the mole ratio between Fe2O3 and Fe is also 1:1.
Therefore, the moles of Fe produced will be the same as the moles of Fe2O3 used.
3) Mole-Mass Conversion:
a) Convert moles of Fe to mass:
Mass of Fe = moles of Fe x molar mass of Fe
Now let's calculate the answers:
1) From step 1b, we found that the moles of Fe2O3 is:
Moles of Fe2O3 = 100.0 g / 159.69 g/mol
Therefore, the mass of solid iron produced will be equal to:
Mass of Fe = Moles of Fe2O3 x Molar mass of Fe
2) a) To find the mass of the excess reactant, we need to calculate the moles of AgNO3 and CaCl2:
Moles of AgNO3 = 7.000 mol
Moles of CaCl2 = 7.000 mol
Since the mole ratio between AgNO3 and CaCl2 in the balanced chemical equation is 2:1:
AgNO3 is the excess reactant, and CaCl2 is the limiting reactant.
b) The mass of the limiting reactant (CaCl2) can be calculated as:
Mass of CaCl2 = Moles of CaCl2 x Molar mass of CaCl2
c) Using the mole ratio from the balanced chemical equation, we can determine the moles of AgCl formed. The mole ratio is 2:2, so it's a 1:1 ratio between the moles of AgNO3 and AgCl.
The mass of each product can be calculated as:
Mass of AgCl = Moles of AgCl x Molar mass of AgCl
Mass of Ca(NO3)2 = Moles of Ca(NO3)2 x Molar mass of Ca(NO3)2
Note: To calculate the molar mass of each compound, look up the individual atomic masses of the elements present and sum them up.
By following these steps and performing the necessary calculations, you will be able to find the answers to the questions asked.