How many grams of solid product can be produced by mixing 45.6 grams of chlorine gas with excess sodium metal?
first, write the balanced equation:
Cl2 (g) + 2Na (s) --> 2NaCl (s)
*chlorine gas is diatomic*
then, do dimensional analysis to cancel units:
g of NaCl = (45.6 g Cl2)*(1 mol Cl2/35.45*2 g Cl2)*(2 mol NaCl/1 mol Cl2)*(58.45 g NaCl/1 mol NaCl)
*35.45 g is atomic mass of Cl, but since it is diatomic, it should be 35.45*2 g Cl2
*23 g is atomic weight of Na, thus 23 + 35.45 = 58.45 g is weight of NaCl
solving this, we obtain 75.2 g NaCl
(always check the number of significant digits)
so there,, :)
To determine the number of grams of solid product that can be produced, we need to balance the chemical equation for the reaction between chlorine gas (Cl2) and sodium metal (Na), and then use stoichiometry to calculate the mass.
The balanced chemical equation for the reaction is:
2 Na + Cl2 -> 2 NaCl
From the balanced equation, we can see that 2 moles of sodium (Na) react with 1 mole of chlorine gas (Cl2) to produce 2 moles of sodium chloride (NaCl).
To calculate the mass of the solid product (NaCl) formed, we need to convert the mass of chlorine gas (Cl2) to moles:
1 mole of Cl2 has a molar mass of approximately 70.9 g (35.45 g/mol × 2).
We can use the molar mass to convert the mass of chlorine gas (Cl2) to moles:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 45.6 g / 70.9 g/mol ≈ 0.6436 mol
According to the stoichiometry of the reaction, 1 mole of Cl2 reacts to produce 2 moles of NaCl. Therefore, 0.6436 moles of Cl2 will produce 2 × 0.6436 = 1.2872 moles of NaCl.
Finally, we can use the molar mass of NaCl (approximately 58.5 g/mol) to convert the moles of NaCl to grams:
mass of NaCl = moles of NaCl × molar mass of NaCl
mass of NaCl = 1.2872 mol × 58.5 g/mol ≈ 75.2772 g
So, approximately 75.28 grams of solid product (NaCl) can be produced by mixing 45.6 grams of chlorine gas with excess sodium metal.