Question

2. Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP, is needed to complete the reaction with 6.25 g of sodium metal. ____Na(s) + ____Cl₂(g) → _____NaCl(s)
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Answer 1

Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)

when we have STP conditions, we can use this conversion: 1 mol = 22.4 L

first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation. 

molar mass of Na= 23.0 g/mol
 ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)

calculations:

$6.25 g Na ( \frac{1 mol Na}{23.0 g} ) ( \frac{1 mol Cl_2}{2 mol Na} ) ( \frac{22.4 L}{1 mol} )= 3.04 L$