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)?

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!

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