Question

1. (2 pts) Write the balanced molecular equation for the reaction that occurs when solutions of HNO3(aq) and Sr(OH)2(aq) are mixed. In order to obtain full credit, you must include state (aq, s, l, or g) of each reactant and product.

Answer:________________________________________

2. (6.5 pts) A titration is performed to find the concentration of 15.00 mL of HNO3 and the end-point is obtained when 35.10 mL of 0.4500 M Sr(OH)2 are used. Find [HNO3]. Show work clearly, circle answer, and provide units for full credit.

3. (6.5 pts) 71.0 mL of NH3 at 155 °C and 810. mmHg react with excess oxygen according to the reaction below. What is the volume of H2O formed when measured at 245 °C and 745 mmHg? Show work clearly, circle answer, and provide units for full credit.

4 NH3(g)+ 3O2(g)→ 2N2(g) + 6H2O (g)

Please provide the whole work process on how to do it on all three questions thanks.

Answer 1

1. 2HNO3(aq) + Sr(OH)2(aq) --------> Sr(NO3)2(aq) + 2H2O(l)

2.

2HNO3(aq) + Sr(OH)2(aq) --------> Sr(NO3)2(aq) + 2H2O(l)

2 moles ------1 moles

HNO3 -------------------------------- Sr(OH)2

M1 =    ------------------------------M2 = 0.45M

V1 = 15ml ------------------------ V2   = 35.1ml

n1   = 2 ----------------------------- n2 = 1

         M1V1/n1     =         M2V2/n2

          M1             =          M2V2n1/n2V1

                           =          0.45*35.1*2/(1*15)

                            = 2.106M

[HNO3]   = 2.106M

3.

P   = 810/760    = 1.0657atm

V   = 71ml = 0.071L

T   = 155+273   = 428K

PV   = nRT

n    = pV/RT

       = 1.0657*0.071/(0.0821*428)

        = 0.002153moles

4 NH3(g)+ 3O2(g)→ 2N2(g) + 6H2O (g)

4 moles of NH3 react with excess of O2 to gives 6 moles of H2O

0.002153 moles of NH3 react with excess of O2 to gives = 6*0.002153/4    = 0.0032295moles of H2O

P = 745/760   = 0.98atm

T   = 245+273    = 518K

n   = 0.0032295moles

PV = nRT

V   = nRT/P

     = 0.0032295*0.0821*518/0.98    = 0.14L

0.14L of H2O >>>>answer