How many grams of Na2CO3 are required to completely react with 35.6mL of a 0.315M HCl solution? Na2O3(s) + 2HCl(aq) --> 2NaCl(aq( + CO2(g) + H2O(I)

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To solve this problem, we need to use the concept of molarity and stoichiometry.

Step 1: Write the balanced chemical equation:
2Na2CO3(s) + 2HCl(aq) --> 2NaCl(aq) + CO2(g) + H2O(l)

According to the balanced equation, we see that 2 moles of Na2CO3 react with 2 moles of HCl.

Step 2: Calculate the number of moles of HCl:
Molarity (M) = moles (n) / volume (V)
n = M x V
n = 0.315 M x 0.0356 L = 0.01122 moles

From the balanced equation, we can see that 2 moles of HCl react with 2 moles of Na2CO3. Therefore, the moles of Na2CO3 required will also be 0.01122 moles.

Step 3: Convert moles to grams:
To convert moles to grams, we need the molar mass of Na2CO3. The molar mass of Na is 22.99 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol.
Molar mass of Na2CO3 = (2 × 22.99 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol) = 105.99 g/mol

Grams (m) = moles (n) × molar mass (M)
m = 0.01122 moles × 105.99 g/mol = 1.190078 g

Therefore, approximately 1.19 grams of Na2CO3 are required to completely react with 35.6 mL of a 0.315 M HCl solution.