Question

1) Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation: 3N2H4(l)→4NH3(g)+N2(g)

A) 35.3 g N2H4 - Express your answer using three significant figures.

B) 14.9 kg N2H4 - Express your answer using three significant figures.

2) For each reaction, calculate the mass of the product that forms when 12.5 g of the reactant in red completely reacts. Assume that there is more than enough of the other reactant.

a) 2K(s)+Cl2(g)→2KCl(s) b) 2K(s)+Br2(l)→2KBr(s) c) 4Cr(s)+3O2(g)→2Cr2O3(s) d) 2Sr(s)+O2(g)→2SrO(s)

3) Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+NO(g)

Suppose that 3.7 mol NO2 and 1.2 mol H2O combine and react completely. Which reactant is in excess? Express your answer as a chemical formula.

4) Calculate the molarity of each of the following solutions:

A) 0.19 mol of LiNO3 in 5.48 L of solution - Express your answer using two significant figures. B) 62.1 g C2H6O in 2.18 L of solution - Express your answer using three significant figures.

C) 16.14 mg KI in 106 mL of solution - Express your answer using three significant figures.

Thank you so much in advance!

Answer 1

3N₂H₄(l) → 4NH₃(g) + N₂(g)

Molar mass of N₂H₄ = 32 g/mole

Thus, moles of N₂H₄ in 35.3 g of it = mass/molar mass = 35.3/32 = 1.103

moles of N₂H₄ in 14900 g of it = mass/molar mass = 14900/32 = 465.625

A) 35.3 g N₂H₄

Thus, moles of NH₃ formed = (4/3)*moles of N₂H₄ reacting = (4/3)*1.103 = 1.471

4) Molarity = moles of solute/volume of solution in litres

A) 0.19 mol of LiNO3 in 5.48 L of solution

Molarity = 0.19/5.48 = 0.035 M

B) 62.1 g C2H6O in 2.18 L of solution

Molar mass of C2H6O = 46 g/mole

Thus, moles of C2H6O = mass/molar mass = 62.1/46 = 1.35

Molarity = 1.35/2.18 = 0.619 M

C) 16.14 mg KI in 106 mL of solution

Molar mass of KI = 166 g/mole

Thus, moles of KI in 0.01614 g of it = mss/molar mass = 0.01614/166 = 9.723*10^-5

Molarity = (9.723*10^-5)/0.106 = 9.173*10^-4 M

3) 3NO2(g)+H2O(l)→2HNO3(l)+NO(g)

As per the balanced reaction , NO2 & H2O reacts in the molarratio of 3:1

Thus, for 1.2 moles of H2O , moles of NO2 required = 3.6

Clearly, NO2 is in excess and the excess amount is 0.1 mole

2) There is no reactant marke red.Please repost Q2.

B) 14.9 kg N2H4 -

Thus, moles of NH₃ formed = (4/3)*moles of N₂H₄ reacting = (4/3)*465.625 = 620.834