suppose 5.00g of sodium carbonate reacts completely with excess of hydrochloric acid. Show how to calculate the mass (g) of sodium chloride that would be formed.

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

What is the limiting reagent?

how many moles of sodium carbonate is in 5 grams? From the balanced equation, one gets twice that number of moles of sodium chloride. That number can be converted to grams of sodium chloride.

To determine the limiting reagent, you need to compare the stoichiometry of the reactants. In this case, we have 5.00g of sodium carbonate and an excess of hydrochloric acid. The balanced equation tells us that 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid to produce 2 moles of sodium chloride.

First, we need to calculate the number of moles of sodium carbonate present in 5.00g. The molar mass of sodium carbonate (Na2CO3) is:
2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol

Using the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass, we can find the number of moles:
n = 5.00 g / 105.99 g/mol ≈ 0.0471 mol

According to the balanced equation, for every 1 mole of Na2CO3, we expect 2 moles of NaCl to be formed. Therefore, if sodium carbonate is the limiting reagent, we would expect to get:
2 * 0.0471 mol = 0.0942 mol of NaCl

Since the stoichiometry of the balanced equation shows that the ratio is 1:2 between Na2CO3 and NaCl, this means that 0.0942 mol of sodium chloride would be formed.

To calculate the mass of sodium chloride, we will use the formula:
mass = n * M, where mass is the mass in grams, n is the number of moles, and M is the molar mass.

The molar mass of sodium chloride (NaCl) is:
22.99 g/mol + 35.45 g/mol = 58.44 g/mol

Using this molar mass to calculate the mass of sodium chloride:
mass = 0.0942 mol * 58.44 g/mol ≈ 5.50 g

Therefore, the mass of sodium chloride that would be formed is approximately 5.50 grams.