What volume in L of 0.27 M Ca(OH)2 is required neutralize 0.40 moles of HCI? The

Question

What volume in L of 0.27 M Ca(OH)2 is required neutralize 0.40 moles of HCI? The balanced reaction is: 2HCl(aq) + Ca(OH)2(aq) 2H2O(I) + CaCl2(aq) Using Balanced Equations II Answer What volume in L of chlorine at 30°C and 719 torr is required to react completely with 9.84 g aluminum 2Al(s) + 3C12(g) 2AlCl3(s) Calculate the moles chlorine needed (as in the previous problem), then use the ideal gas law--PV nRT. Additional help is given in the feedback. Answer: Procedure: g A --> mol Al--> mol Clg. Next solve for V in PV = nRT. Did you remember to convert torr to atm and C to K? The following reaction takes place at a certain elevated temperature: Fe2O3(s) + 3CO(g) 2Fe() + 3CO2(s) What is the percent yield of iron if 57.7 g Fe203 in excess CO produces 21.4 g Fe? The M.W. of Fe2O3 is 159.7 g/mol and the M.W. of CO is 28.01 g/mol. Recall that the percent yield is the actual yield divided by the theoretical yield times 100%. Additional help is given in the feedback. Answer:

Explanation / Answer

Ans 1

From the stoichiometry of the reaction

2 moles of HCl required = 1 mol of Ca(OH)2

1 moles of HCl required = 0.5 mol of Ca(OH)2

0.40 moles of HCl required = 0.40 x 0.5 = 0.2 mol of Ca(OH)2

Volume of Ca(OH)2 = moles/molarity

= 0.2 moles / 0.27 moles/L

= 0.741 L

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