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When 2.5g of Al reacts with 5.0g of Cl2, 8.3g of AlCl3 is formed. Calculate the limiting reactants, theoretical yield, and percent yield for the reaction. 2Al + 3Cl3 ---> 2AlCl3
2Al + 3Cl2 ==> 2AlCl3Find limiting reactant:moles Al present = 12.8 g Al x 1 mol Al/27.0 g = 0.474 moles Al (÷2 ->0.237)moles Cl2 present = 31.9 g Cl2...
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HCl solution consists of 28% HCl by mass and has a density of 1.14g/ml. Find volume of the solution required to react with 1.87g Al as in the reaction:
2Al + 6HCl = 2AlCl3 + 3H2
2Al + 6HCl = 2AlCl3 + 3H2M (Al) = 26.98 g/mol2 x 26.98 reactsM (HCl) = 36.46 g/mol6 x 36.46 g reactsGiven mass of Al: 1.87 gThe ratio is:1.87 /...
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Based on the following equation:
6HCL(aq)+ 2AL(s)> 2ALCL3(aq) + 3H2 (g)
What mass of aluminum is needed to produce 500 mL of hydrogen gas at 25 degrees Celcius and 100 kPa?
P = 100 kPa T = 25oC + 273 = 298 KV = 500 mL = 0.5 LR = 8.314 kPa L K-1 mol-1Solution:The balanced chemical equation:6HCl(aq) + 2Al(s) =...
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What it the mass concentration of the excess reagent in the non Stoichemtric reaction (6HCl+2Al ➡2AlCl3+3H2)
The reactant that produces a lesser amount of product is the limiting reagent.The reactant that produces a larger amount of product is the excess reagent.To find the amount of remaining...
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what is the expected volume of 10 grams of oxygen gas at STP? given that the volume of 1 mole of a gas at STP is 22.4 Liters
M (O2)=32 /moln=m/Mn (O2)=10/32=0.3 molV (O2) = 0.3 x 22.4 = 6.72 l
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When aluminum reacts with hydrochloric acid, HCl, both aluminum chloride and hydrogen gas are produced. If 108 grams of aluminum is reacted, what volume of gas may be collected at STP?
Solution:Balanced chemical equation:2Al + 6HCl → 2AlCl3 + 3H2According to the equation above: n(Al)/2 = n(H2)/3The molar mass of Al is 27 g/mol.Hence,Moles of Al = (108 g Al) ×...
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for CH4(g) + 3Cl2(g) ——> CHCl3(g) + 3HCl(g) how many grams of CH4 is needed to produce 45.5 g CHCl3
Solution:The molar mass of CH4 is 16.04 g/molThe molar mass of CHCl3 is 119.38 g/molCalculate the moles of CHCl3:(45.5 g CHCl3) × (1 mol CHCl3 / 119.38 g CHCl3) =...
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How many grams of aluminum chloride could be produced from 34.0g of aluminum and 39.0g of chlorine gas? How many moles of the excess reagent are left?
48.9 g of aluminium chloride is formed1.07 moles of aluminium is excess.
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For the reaction shown, find the limiting reactant for each of the following initial amount of reactants.
2Al (s)+3Cl2 (g) —> 2AlCl3 (s)
1.0g Al; 1.0g Cl2
M(Al) = 26.98 g/mol;n=m/M;n(Al) = 1/26.98 = 0.037 moles;M(Cl2) = 70.90 g/mol;n(Cl2) = 1/70.90 = 0.014 moles;The mole ratio Al:Cl2 according to the reaction is 2:3;The mole ratio Al:Cl2 according...
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How many grams of chlorine gas must react to give 3.72 g of BiCl3?
2Bi(s)+3Cl2(g)→2BiCl3(s)
Express your answer with the appropriate units.
2 BiCl3 = 3 Cl22(315.34g) BiCl3 = 3(70.9 g) Cl21g BiCl3 = 3 ×70.9 /(2×315.34)g Cl23.72g = (3 ×70.9 × 3.72) /(2 ×315.34g) Cl2= 1.25g
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