1) Calculate how many moles of NH3 form when each quantity of reactant completel

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!

Explanation / Answer

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

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