I missed the lecture about this lab and don\'t know how to do any of these. I wo

Question

I missed the lecture about this lab and don't know how to do any of these. I would like explanations as well as answers, thanks! Chemistry 151 Questions: 1. In Part 1 of this experiment, is it theoretically possible for the heat gain by cool water to Modern Experimental Chemistry exceed heat loss by warm water? Explain how your answer relates to line 6, heat gained by the calorimeter. 2. (a) Write out the balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide, using smallest whole number stoichiometric coefficients. (b) Write out the net ionie equation for the reaction between hydrochlorie acid and sodium hydroxide. (c) Using your results, what is the ratio ofAHndHSO.) to 11ndHcD2 what should this ratio be? Explain. 3. If in Part 2, you mixed (carefully measured) 30.0 mL of 0.80 M NaOH with 70.0 mL of 0.50 M HCI, which of the two reagents is the limiting reagent for heat of reaction? How many moles of water would be produced? Show Show calculations. 4. What would be the approximate AT obtained after mixing 70.0 mL 1.0 M HCl with 30.0 ml I M NaOH? Use your data. Show calculations. 37 Exp 6 Thermochemistry

Explanation / Answer

1) The question asks me to refer to line 6 of the theory/procedure. You need to upload the experiment so that I can refer to line 6.

2a) Write out the equation taking place between sulfuric acid and sodium hydroxide as

H2SO4 + NaOH ------> Na2SO4 + H2O

Balance Na on both side: multiply NaOH by 2. Therefore,

H2SO4 + 2 NaOH ------> Na2SO4 + H2O

Next balance H on both sides: the left side has 4 H atoms while there are 2 on the right. Multiply H2O on the right by 2.

H2SO4 + 2 NaOH -----> Na2SO4 + 2 H2O

Check that all the atoms are balanced now. Finally write down the state of occurrence of each reactant and product to get the balanced equation as

H2SO4 (aq) + 2 NaOH (aq) ------> Na2SO4 (aq) + 2 H2O (l) (ans)

b) Write down the molecular reaction first:

HCl (aq) + NaOH (aq) -----> NaCl (aq) + H2O (l)

Since we have aqueous solutions of two electrolytes, both will dissociate into ions. Decompose the electrolytes into ions and write

H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) ------> Na+ (aq) + Cl- (aq) + H2O (l)

Cancel the common ions on both sides to write the balanced net ionic equation as

H+ (aq) + OH- (aq) ------> H2O (l)

c) Let H1 be the heat of reaction in (2a) and H2 be the heat of reaction in (2b).

The heat of reaction per mole of water is given as Hrxn (H2SO4) = ½*H1 and Hrxn (HCl) = H2.

The question asks me to use your results; you haven’t uploaded any data and hence I cannot calculate the results.

However, I believe the ratio will be 1:1, since we are calculating the heat liberated for 2 moles of water in (2a) and for 1 mole in (2b). The heat of reaction per mole should be same (since we divide 2a by 2 and 2b by 1).

3) Write out the balanced chemical reaction:

HCl + NaOH ------> NaCl + H2O

Moles NaOH = (volume of HCl in L)*(concentration in moles/L) = (30 mL)*(1 L/1000 mL)*(0.80 moles/L) = 0.024 moles.

Moles HCl = (70.0 mL)*(1 L/1000 mL)*(0.50 moles/L) = 0.035 moles.

Since we have fewer moles of NaOH and as per the balanced equation, NaOH and HCl react on a 1:1 molar ratio, therefore, NaOH will be the limiting reactant (ans).

Again, look at the balanced equation above. The molar ratio of reaction between NaOH and H2O is 1:1. Therefore moles of water produced = moles of NaOH reacted = 0.024 moles (ans).

4) No data provided.

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