Given a #20*g# mass of calcium carbonate, how much carbon dioxide will be generated, when it is added to a #100*mL# volume of #"20% w/w"# solution of #HCl(aq)#?
The units of concentration of the hydrochloric acid are unfortunate.....I gets approx. a
We interrogate the reaction......
How much acid is in excess?
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The balanced chemical equation for the reaction is:
[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) ]
Using the molar mass of CaCO3 (100.09 g/mol), the moles of CaCO3 can be calculated as ( \frac{20 \text{ g}}{100.09 \text{ g/mol}} ).
According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the moles of CO2 produced will be the same as the moles of CaCO3.
The volume of CO2 at standard temperature and pressure (STP) can be calculated using the ideal gas law: ( V = n \times 22.4 \text{ L/mol} ), where ( n ) is the moles of CO2.
Finally, the volume can be converted from liters to milliliters ((1 \text{ L} = 1000 \text{ mL})).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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