What mass (g) of Na2CO3 would be needed to react with 0.250 moles of HCI according to the equation
Na2CO3 + 2HCI ---> 2NaCI + CO2 + H2O.
My answer is 13.2g but I'm not sure if it is correct. Can someone please verify this for me. Any help is very much appreciated. Thanks
each mole of Na2CO3 reacts with 2 moles of HCl.
0.250/2 = 0.125 moles Na2CO3 = 13.25g
So, you are correct.
Thanks for checking
To find the mass of Na2CO3 needed to react with a given amount of HCl, you need to use stoichiometry. Below are the steps to calculate the mass:
1. Start with the number of moles of HCl given in the problem: 0.250 moles.
2. Use the balanced chemical equation to determine the mole ratio between HCl and Na2CO3. According to the equation, it is 2 moles of HCl : 1 mole of Na2CO3.
3. Set up a proportion to find the number of moles of Na2CO3:
(0.250 moles HCl) / (2 moles HCl) = (x moles Na2CO3) / (1 mole Na2CO3).
Solving for x gives:
x = (0.250 moles HCl) * (1 mole Na2CO3) / (2 moles HCl)
= 0.125 moles Na2CO3.
4. Now, convert the moles of Na2CO3 into grams. To do this, you need to know the molar mass of Na2CO3. The molar mass of Na is approximately 23 g/mol, C is approximately 12 g/mol, and O is approximately 16 g/mol (each O is counted twice in Na2CO3). Adding them together, you get 106 g/mol as the molar mass of Na2CO3.
5. Calculate the mass of Na2CO3:
mass = (0.125 moles Na2CO3) * (106 g/mol Na2CO3)
= 13.3 g Na2CO3 (rounded to one decimal place).
Therefore, the correct mass of Na2CO3 needed to react with 0.250 moles of HCl is approximately 13.3 grams, not 13.2 grams as you initially calculated.
Please note that the number of significant figures has been taken into account throughout the calculation, and the final answer has been rounded accordingly to one decimal place.