The average rate of the following reaction at a certain temperature is 9.04×10 -

Question

The average rate of the following reaction at a certain temperature is 9.04×10-1 M/min. Calculate the value for the rate of change of [HNO3], in M/min, at this temperature. Report your answer to three significant figures in scientific notation.

H2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2HNO3(aq)

If the intial concentration of H2SO4 is 0.684 M and the initial concentration of Sr(NO3)2 is 0.562 M, determine the concentration of H2SO4 (in M) after 34 seconds. Assume the volume of the solution does not change and report your answer to three significant figures in scientific notation.

Explanation / Answer

H2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2HNO3(aq)

1     :               1                 : 1          :   2

9.04*10^-1                                        : 18.08 *10^ -1

Therefore chane in conc of HNO3 = 1.81 M/min

Moles changed of H2SO4 in one min   = 9.04 * 10^ -1

      "                               in one sec   = ( 9.04 * 10^ -1) / 60

      "                               inn 34 sec   =   9.04*10^ -1 * 34 / 60    =.512 M

H2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2HNO3(aq)

1     :               1                 : 1          :   2

.512 M           .512   M             .512 M       1.024 M

So conc of H2SO4 after 34 seconds = .684-.512   =   0.172 M

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