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4 moles
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
58.9g
8 g of magnesium chloride is obtained.
Balanced equation and potassium limits and drives the reaction.2K + Cl2 -> 2KCl6.75 grams K (1 mole K/39.10 grams)(2 mole KCl/2 mole K)(74.55 grams /1 mole KCl)= 12.9 grams potassium chloride produced==============================
4 moles
1.26 mol of AlCl3
Four:2 Al + 3 Cl2 --> 2 AlCl3so: 4 Al + 6 Cl2 --> 4 AlCl3
4 moles
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
We need 6,935 g aluminium; the excess of aluminium is 0,97 g.
Sodium has an atomic weight of 22.99 g/mol. Chlorine has an atomic weight of 35.45 g/mol. NaCl has an atomic weight of 58.44 g/mol. Therefore 92g of sodium would yield 233.86g of sodium chloride (NaCl).
The balanced chemical equation for the reaction between calcium and chlorine gas to produce calcium chloride is: Ca + Cl2 -> CaCl2. From this equation, we can see that one mole of calcium reacts with one mole of chlorine gas to produce one mole of calcium chloride. The molar mass of calcium is 40.08 g/mol and the molar mass of chlorine gas is 70.90 g/mol. This means that 10.0 grams of calcium is equivalent to 0.249 moles of calcium and 20.0 grams of chlorine gas is equivalent to 0.282 moles of chlorine gas. Since the ratio of calcium to chlorine gas in the balanced chemical equation is 1:1, this means that 0.249 moles of calcium would react completely with 0.249 moles of chlorine gas, leaving an excess of 0.033 moles (or 2.34 grams) of chlorine gas. The limiting reactant in this reaction is calcium, and the maximum amount of calcium chloride that can be produced is equivalent to the number of moles of the limiting reactant, which is 0.249 moles (or 27.8 grams) of calcium chloride.
58.9g
The answer is: approx. 327 g.
384.5g