Question

When 107 g NaCl (MM=58.44 g/mol) and 135 g H2SO4 (MM=98.08 g/mol) are mixed and react according to the equation below, what is the mass in grams (g) of Na2SO4 produced? Give your answer rounded to a whole number. Do not enter units. (Hint: Find the limiting reactant; MM Na2SO4=142.04 g/mol) 2NaCl H2SO4→ Na2SO4 2HCl

Answer 1

Answer: 129.9666 grams of Na2SO4

Step-by-step explanation:

We must use limiting reactants in this problem.

First, let's calculate the number of moles of NaCl, which is
`(107)/(58.44)` or 1.83 moles. Now, let's calculate the number of moles of H2SO4, which is
`(135)/(98.08)` or 1.376 moles. Now, since we know that the ratio of H2SO4 used to NaCl used is 1/2, let's multiply 1.376 by 2 to get 2.752 moles.

Since 1.83 is less than 2.752, we can conclude that NaCl is the limiting reactant.

Since the ratio of NaCl used to Na2SO4 formed is 2/1, we need to divide 1.83 by 2 (
`(1.83)/(2)`) or 0.915 moles.

Multiplying 0.915 moles by the molar mass of Na2SO4 (142.04) we arrive at the answer.

129.9666 grams of Na2SO4 is formed