Question

9. How many moles of NO2 are contained in 45.1g of NO2?

10. How many grams of Al are reuired to react completely with 4.30g of HCi according to the following equation? 2Al +6HCl -> 2AlCl3 + 3H2

11. Concentrated sodium hydoxide solution contains 50.0% by mass NaOH an has density of 1.52g/mL. What volume of this solution in mL. required to abotain 1.35g of N6OH?

Answer 1

Solution :-

9. How many moles of NO2 are contained in 45.1g of NO2?

Solution :- formula to calculate the moles is

Moles = mass in gram / molar mass

Molar mass of NO2 = 46.0055 g per mol

Moles of NO2 = 45.1 g / 46.0055 g per mol

                          = 0.980 mol NO2

10. How many grams of Al are reuired to react completely with 4.30g of HCi according to the following equation? 2Al +6HCl -> 2AlCl3 + 3H2

Solution :- using the mole ratio of the Al and HCl we can calculate the mass of Al needed to react with 4.30 g HCl

(4.30 g HCl * 1 mol / 36.5 g)*(2 mol Al / 6 mol HCl)*(26.982 g/ 1 mol ) = 1.06 g Al

Soo the mass of Al needed for the complete reaction is 1.06 g Al

11. Concentrated sodium hydoxide solution contains 50.0% by mass NaOH an has density of 1.52g/mL. What volume of this solution in mL. required to abotain 1.35g of N6OH?

Solution : -

50 % by mass solution means 50 g solute and 50 g water

So

Lets find the mass of solution to get 1.35 g NaOH

1.35 g NaOH * 100 % / 50 % = 2.70 g solution

Now lets convert the gram to ml

Volume = mass/ density

               = 2.70g / 1.52 g/ml

               = 1.78 ml

So we need to use 1.78 ml solution to get 1.35 g NaOH