Question

For the reaction 2Na(s) + Cl_2(g) = 2NaCl(s),

how many grams of NaCl could be produced from 103.0 g of Na and 13.0 L of Cl_2 (at STP)?

Answer 1

2 Na(s) + Cl₂(g) = 2 NaCl(s) 22,4 L in STP

Number of moles :

n = m / molar mass

Na = 103.0 / 22.98

Na = 4.48 moles

Cl₂ ( is limiting reagent )

13 L / 22,4 L = 0.580 moles Cl₂

2 ( Sodium stoichiometric coefficient)* 0.580 = 1.16 mol NaCl

Molar mass NaCl = 58.44 g/mol

1.16 * 58.44 = 67.7904 g of NaCl

hope this helps!