510 g Al2S3 is equal to 3,396 moles.
100/150.158 is 0.666 moles
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}).
The number of moles is mass in g/molar mass in g.
The number of moles 9,92.10e-5.
The number of moles in 432 g Mg (OH)2 is 7,407.
100/150.158 is 0.666 moles
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 )
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
The number of moles is mass in g/molar mass in g.
The formula is: number of moles = g Be/9,012.
The number of moles 9,92.10e-5.
0.0027 moles.
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.
22.99 g of C28H44O is equal to 0,058 moles.
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.
The number of moles in 432 g Mg (OH)2 is 7,407.
7.24 moles.