10. What are the most likely products of the following reaction (you don\'t need
ID: 949132 • Letter: 1
Question
10. What are the most likely products of the following reaction (you don't need to balance the reaction, just predict the products)? I. Iron reacts with oxygen to form iron (III) oxide, as shown in the unbalanced reaction below. According to mole(s) of the balanced equation, one mole of iron (IIl) oxide is stoichiometrically equivalent toX iron. (fill in the blank) Balonce 2) (a) one (b) two (c) three (d) four (e) five 2. Refer to the reaction in question 1. How many grams of iron (II) oxide will be formed from 125 g of iron? (a) 59.8 g (b) 89.7 g (e)179 g (d) 247 g (e) 359 g 3. Refer to the reaction in question 1. How many grams of oxygen are needed to produce 10.5 g of solid iron (III) oxide (a) 2.11 g (b) 1.40 g (c) 3.16 g (d) 5.25 g (e) 10.5 g Questions 4 and 5 pertain to the following reaction: N;(g) + 3H2(g) 2NH3(g) 4. In one experiment, 5.00 g of hydrogen was treated with 10.0 g of nitrogen. Which of the following is true for this experiment? (a) the theoretical yield of ammonia was 12.2 g, and 2.84 g of hydrogen was unreacted (left over) (b) the theoretical yield of ammonia was 28.2 g, and 3.15 g of nitrogen was unreacted (left over) (c) the theoretical yield of ammonia was 6.10 g, and 2.16 g of hydrogen was unreacted (left over) (d) the theoretical yield of ammonia was 42.2 g, and 2.16 g of nitrogen was unreacted (left over) (e) the theoretical yield of ammonia was 6.10 g, and 3.15 g of hydrogen was unreacted (left over) s. In a different experiment, 4.50 x 10 g of hydrogen reacted with excess nitrogen to produce 1575 g of ammonia. What was the percent yield of the reaction? (b) 30.8% (a) 20.7% (c) 41.5% (d) 618% (c) 73.6% 6. Hydrochloric acid is widely used as a laboratory reagent. Calculate the number of moles of HCI in 62.85 .o265L c) 0.139 mol (d)0285 mol mL of 0.453 Mf hydrochloric acid 0G285 a) 28.5 mol b) 1.04 mol e)0.00721 mol 7. What will be the final volume of a solution prepared by diluting 35.0 mL of 6.50 M sodium hydroxide to a concentration of 3.75 M a) 60.7 mL b) 351 mL c) 4.20 mL d) 0.577 mL e) 3.75 L 60.7mL b) 3S1 mL, 6) 420 ml. ) 0.577 mL ) 3.75 LExplanation / Answer
Answer – 1) In this one we are given the unbalanced equation-
Fe(s) + O2 ---> Fe2O3
In this reaction in the reactant side there are 1 Fe and 2 O element and in the product side there are 2 FE and 3 O. So first we need to balance the Fe, so we need to put 2 in front of Fe, so reaction is as follow –
2 Fe(s) + O2 ---> Fe2O3
Now from both side Fe gets balanced now we need to balance the O from both side and for that we need to put 3/2 in front of O2 so it make 3 O in both side as follow –
2 Fe(s) + 3/2 O2 ---> Fe2O3
Now we know the mole coefficient is not in fraction number, so for making the whole number we need to multiply whole equation by 2 as follow –
4 Fe(s) + 3 O2 ---> 2 Fe2O3
Now this equation gets balanced from both side and the for one moles of Fe2O3 is stoichiometrically equivalent to 2 moles of Fe, since
2 moles of Fe2O3 = 4 moles of Fe
So, 1 moles of Fe2O3 = ?
= 2 moles of moles of Fe
2) Given, mass of Fe = 125 g , mass of Fe2O3 = ?
First we need to calculate the moles of Fe
We know, moles = given mass / molar mass
Moles of Fe = 125 g / 55.845 g.mol-1
= 2.24 moles
From the balanced equation
4 moles of Fe = 2 moles of Fe2O3
So, 2.24 moles of Fe = ?
= 1.12 moles of Fe2O3
Mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
= 1.12 moles of Fe2O3 * 159.68 g/mol
= 179 g of Fe2O3
So answer for this one is option C) 179 g
3) Given, mass of Fe2O3 = 10.5 g , mass of O2 = ?
First we need to calculate the moles of Fe2O3
We know, moles = given mass / molar mass
Moles of Fe2O3 = 10.5 g / 159.68 g.mol-1
= 0.00658 moles
From the balanced equation
2 moles of Fe2O3 = 3 moles of O2
So, 0.00658 moles of Fe2O3 = ?
= 0.0986 moles of O2
Mass of O2= moles of O2 * molar mass of O2
= 0.0986 moles of O2* 32.0 g/mol
= 3.16 g of O2
So answer for this one is option C) 3.16 g
4) Given, reaction – N2(g) + 3H2(g) -----> 2NH3(g)
Mass of H2 = 5.00 g , mass of N2 = 10.0 g
First we need to calculate moles of each
Moles of H2 = 5.00 g / 2.0158 g.mol-1
= 2.48 moles of H2
Moles of N2 = 10.0 g / 28.014 g.mol-1
= 0.357 of N2
Now we need to calculate the limiting reactant –
Moles of NH3 from H2
From the balanced equation –
3 moles of H2 = 2 moles of NH3
So, 2.48 moles of H2 = ?
= 1.65 moles of NH3
Moles of NH3 from N2
From the balanced equation –
1 moles of N2 = 2 moles of NH3
So, 0.357 moles of N2 = ?
= 0.714 moles of NH3
So moles of NH3 is lowest form the N2 , so limiting reactant is N2
And moles of NH3 = 0.714 moles
Mass of NH3 formed = 0.714 moles * 17.0307 g/mol
= 12.16 g of NH3
So moles of H2 reacted –
1 moles of N2 = 3 moles of H2
So, 0.357 moles of N2 = ?
= 1.07 moles of H2
So, moles of H2 unreacted = 2.48 moles – 1.07 moles
= 1.41 mole of H2
So, mass of H2 excess = 1.41 moles * 2.0158 g/mol
= 2.84 g of H2
So option a) the theoretical yield of NH3 is 12.2 g and 2.84 g of hydrogen was unreacted
5) given, mass of H2 = 450 g , excess reactant is N2 , mass of actual NH3 = 1575 g
Moles of H2 = 450 g / 2.0158 g.mol-1
= 223.2 moles of H2
Moles of NH3 from H2
From the balanced equation –
3 moles of H2 = 2 moles of NH3
So, 223.2 moles of H2 = ?
= 148.8 moles of NH3
Mass of NH3 formed = 148.8 moles * 17.0307 g/mol
= 2534.5 g of NH3
So percent yield of NH3 = actual yield / theoretical yield *100 %
= 1575 g / 2534.5 g * 100 %
= 62.1 %
So nearest answer is d) 61.8 %
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