How many grams of HCL needed to react with 4000 g of Na2co3?

Answer 1

First write don the BALANCED reaction eq'n

2HCl(aq) + Na2CO3(aq) = 2NaCl)aq_ + H2O(l) + CO2(g)

Note the molar ratios are 2:1 ::2:1:1

So two moles of HCl reacts with one mole of sodium carbonate .

Next we need to convert 4000 g to moles.

Using the Periodic Table and the Atomic Weights.

2 x Na = 2 x 23 = 46

1 x C = 1 x 12 = 12

3 x O = 3 x 16 = 48

Adding for one molecule(Na2CO3) = 106

Hence moles (Na2CO3) = 4000 / 106 = 37.735849 moles. This is equivalent to 1 molar ratio above.

Hence moles (HCl) = 2 x 37.735849 = 75.471698 . This is equivalent to 2 molar ratios above.

Hence

Mass (HCl) is

75.471698 = mass(g) / 36.5 (The Mr of HCl).

mass (g) = 36.5 X 75.471698 = 2754.71698 g. ~ 2.755 kg.

Answer 2

The balanced chemical equation for this reaction is 2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2. From the equation, 1 mole of Na2CO3 reacts with 2 moles of HCl. Calculate the number of moles of Na2CO3 in 4000g, then use the mole ratio to find the moles of HCl needed. Finally, convert moles of HCl to grams.

Answer 3

Na2CO3 + 2HCl ------> 2NaCl + CO2 + H2O

Using zero decimal places the molecular weight of sodium carbonate is 106 so therefore 4000/106 x 36.5 x 2 = 2755g of pure Hydrogen Chloride.