Multiple Choice
How many grams of sodium metal are needed to generate 4.00 L of hydrogen gas at STP from the reaction with water?
3. Chemical Reactions
Stoichiometry
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What is the concentration of CaCO₃ in mol/L after a 0.2557 g sample is reacted with excess HCl, given that the excess HCl is neutralized by 31.62 mL of NaOH with a concentration of 0.0251 mol/L?
A
0.1007 mol/L
B
0.00255 mol/L
C
0.0502 mol/L
D
0.0251 mol/L
Verified step by step guidance1
Calculate the moles of NaOH used in the reaction by multiplying its volume (in liters) by its concentration. Use the formula: \( \text{moles of NaOH} = \text{volume (L)} \times \text{concentration (mol/L)} \).
Since NaOH neutralizes the excess HCl, the moles of NaOH used are equal to the moles of excess HCl. Therefore, determine the moles of excess HCl using the moles of NaOH calculated in the previous step.
Determine the initial moles of HCl added to the reaction. Since the problem states that the HCl is in excess, the moles of HCl initially added will be greater than the moles of excess HCl.
Calculate the moles of HCl that reacted with CaCO₃ by subtracting the moles of excess HCl from the initial moles of HCl.
Use the stoichiometry of the reaction between CaCO₃ and HCl to find the moles of CaCO₃. The balanced chemical equation is: \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \). From this, determine the moles of CaCO₃ and then calculate its concentration by dividing the moles of CaCO₃ by the volume of the solution in liters.
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