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How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
What is the mass of Al formed when 0.500 moles Al2S3 is reduced completely with excess H2?
To determine the mass of aluminum (Al) formed from the complete reduction of 0.500 moles of aluminum sulfide (Al2S3), we first need to establish the stoichiometry of the reaction. The balanced equation for the reduction of Al2S3 is: [ Al2S3 + 6H2 \rightarrow 2Al + 3H2S. ] From this equation, one mole of Al2S3 produces two moles of Al. Therefore, 0.500 moles of Al2S3 will yield (0.500 \times 2 = 1.000) moles of Al. The molar mass of aluminum is approximately 27.0 g/mol, so the mass of Al produced is (1.000 , \text{mol} \times 27.0 , \text{g/mol} = 27.0 , \text{g}).
How is a number of moles of an element determined from a known mass?
The number of moles is mass in g/molar mass in g.
What is the number of moles in 0.025 g NH42Cr2O7?
The number of moles 9,92.10e-5.
What is the number of moles in 432 g Mg (OH)2?
The number of moles in 432 g Mg (OH)2 is 7,407.
How many moles of Al2S3 are in 100 g of Al2S3?
100/150.158 is 0.666 moles
What is the mass of Al formed when 0.500 moles Al2S3 is reduced completely with excess H2?
To determine the mass of aluminum (Al) formed from the complete reduction of 0.500 moles of aluminum sulfide (Al2S3), we first need to establish the stoichiometry of the reaction. The balanced equation for the reduction of Al2S3 is: [ Al2S3 + 6H2 \rightarrow 2Al + 3H2S. ] From this equation, one mole of Al2S3 produces two moles of Al. Therefore, 0.500 moles of Al2S3 will yield (0.500 \times 2 = 1.000) moles of Al. The molar mass of aluminum is approximately 27.0 g/mol, so the mass of Al produced is (1.000 , \text{mol} \times 27.0 , \text{g/mol} = 27.0 , \text{g}).
How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?
The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )
What is the number of moles of argon in 607 g of argon?
the equation is Xg multiplied by the moles/grams of X = moles of X (the grams cancel leaving you with moles) 607g Ar x 1 mole/ 39.95g = 15.19 moles
How is a number of moles of an element determined from a known mass?
The number of moles is mass in g/molar mass in g.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
What is the number of moles in 0.025 g NH42Cr2O7?
The number of moles 9,92.10e-5.
How many moles of SO2 are present in 250.0 g?
Number of moles is determined by dividing molar mass into the number of grams. SO2 has a molar mass of 64.066 g. To find the number of moles in 250.0 g of SO2, divide 250.0 g by 64.066 g. This gives you just over 3.9 moles.
What is the number of moles in 15 g of As H3?
0.0027 moles.
What is the number of moles if you have 22.99 g of C28H44O?
22.99 g of C28H44O is equal to 0,058 moles.
How many moles are in 4.5 g of AgNO3?
To find the number of moles in 4.5 g of AgNO3, you first need to determine the molar mass of AgNO3 which is 169.87 g/mol. Then you can use the formula: moles = mass / molar mass. Therefore, moles = 4.5 g / 169.87 g/mol ≈ 0.0265 moles.
What is the number of moles in 432 g Mg (OH)2?
The number of moles in 432 g Mg (OH)2 is 7,407.
