Explain how will you determine which of the two reagents,2NO_3PO_4.12H_2O,is 38a

Question

Explain how will you determine which of the two reagents,2NO_3PO_4.12H_2O,is 38aCl_2.2H_2O, is the limiting reagent. A 108.5 g sample of an unknown mixture of NH_3(g) and CuO_(s) was heated at high temperatures and yielded 10.6 g of N_2(g). 2Nh_3(g)+3CuO_(s)+ N_2(g)+3Cu_(s)+3H_2O_(g) The limiting reactant was determined to be CuO_(s).What is the percentage of NH_3(g) and CuO_(s) in the original mixture? (Molar Masses: N=14.0g/mal, H=1.00g/mal, Cu = 63.5g/mol, O=16.00g/mol Why is it important to ensure that your product is dry? An essential step in the industrial production of the element Iodine is the reaction of Sodium Iodate with Sulfur Dioxide: 2NaIO_3+4H_2_O+5SO_2 Na_2SO_4+4H_2SO_4+I_2 What is the mass od So_2 required to completely react with 100g of sodium Iodate? (Molar Masses: I=126.9 g/mol, S=32.10 g/mol, O=16.00 g/mol, Na =23.00 g/mol) List the two reasons why we are using suction filtration in this experiment. After you have completely transferred your precipitate to your Buchner funnel you are required to rinse it with hot water. Why is this step necessary?

Explanation / Answer

Solution :-

Q4). 2NH3 + 3CuO ----- >N2 + 3Cu + 3H2O

Lets find the moles of N2

Moles of N2 = mass / molar mass

                      = 10.6 g N2 / 28.014 g per mol

                     = 0.37838 mol N2

Now lets calculate moles of NH3 using the mole ratio of the N2 and NH3

0.37838 mol N2 * 2 mol NH3 / 1 mol N2 = 0.79676 mol NH3

Now lets calculate the mass of NH3

Mass of NH3 = moles * molar mass

                        = 0.79676 mol * 17.03 g per mol

                       = 13.57 g NH3

So the mass of NH3 = 13.57 g

Lets find the % of NH3

% NH3 = (mass of NH3/ mass of mixture )*100%

             = (13.57 g / 108.5 g)*100%

             = 12.5 % NH3

% of CuO = 100 % - % NH3

                    = 100 % - 12.5 %

                    = 87.5 % CuO

Q5) It is important ensure that the product is dry because if it is not dried well then it contains the moisture in it which increases the mass of the compound. Due to additional mass of the moisture in the product it increases the percent yield than actual yield of the product.

Also the moisture content can decompose the product over the time.

Q6)

2NaIO3   + 4H2O +5SO2 ------ > Na2SO4 +4H2SO4 + I2

What mass of SO2 needed to react with 100 g NaIO3

Using the mole ratio of the SO2 and NaIO3 lets calculate the mass of the SO2 needed

(100 g NaIO3*1 mol / 197.9 g)*(5 mol SO2 / 2 mol NaIO3)*(64.1 g / 1 mol SO2)= 80.97 g SO2

So the mass of SO2 needed is 80.97 g SO2

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.