266,86 g aluminium chloride are obtained.
SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
The answer is: approx. 327 g.
6,36 g of silver chloride are obtained.
The concentration is 50,6 g/L.
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams
36.4 Grams
The formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
The gram atomic mass of aluminum is 26.9815; the gram atomic mass of chlorine is 35.453; and the formula of aluminum chloride is AlCl3, showing that three atoms of chlorine are required for each atom of aluminum in the compound. Therefore, mass ratio of chlorine to aluminum in the compound is [3 X (35.453])/26.9815 or 3.942. The ratio of reactant chlorine stated to be available to reactant aluminum stated to be available is 29.0/24.0 or 1.20, so that chlorine is clearly the limiting reactant. Therefore, the mass of aluminum in the maximum mass of aluminum chloride that can be made from the reactants stated is 29.0/3.942 or about 7.357 grams, and that added to the stated 29.0 g of chlorine constitutes 36.4 grams total of aluminum chloride, to the justified number of significant digits.
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.
75 g sodium chloride contain 29,75 g sodium.
The molar mass of anhydrous aluminum chloride is 133,34 grams.
To find the answer, we multiply the 7 grams of NaCl by the ratio of the molar mass of chlorine over the molar mass of sodium chloride. By doing this, we find that there are about 4.25 grams of chlorine in 7 grams of NaCl.
Lithium chloride is not transformed in calcium chloride.
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.
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================
70
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