3,80 g Zn have 0,058 moles.
The equivalent of 125 g zinc is 1, 91 moles.
3.65(g) / 65.38(g/mol) = 0.0558 (mol)
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
For this you need the atomic mass of Zn. Then take the mass in grams and divide it by the atomic number (multiplied by one mole for units to cancel) to find number of moles. Zinc's atomic mass is 65.4 grams.22.5 g Zn / (65.4 grams) = .344 moles Zn
First, write a balanced equation for this reaction. The reactants are HCl and Zn and the products are ZnCl2(aq), and H2(g). For how to write a balanced equation, see the Related Questions to the left. Then, convert the grams of Zn into moles of Zn. To do that, see the Related Questions to the left. Then use stoichiometry to determine how many moles of HCl are necessary to react with that number of moles of Zn. See the Related Questions to the left for how to solve stoichiometry problems. Finally, determine how many milliliters of solution you need to get that many moles of HCl. To do that, use this equation: number of moles = number of liters * molarity
0,040 moles Zn equal 2,614 g.
The equivalent of 125 g zinc is 1, 91 moles.
0,356 moles of zinc contain 23,27 g.
3.65(g) / 65.38(g/mol) = 0.0558 (mol)
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
For this you need the atomic mass of Zn. Then take the mass in grams and divide it by the atomic number (multiplied by one mole for units to cancel) to find number of moles. Zinc's atomic mass is 65.4 grams.22.5 g Zn / (65.4 grams) = .344 moles Zn
First, write a balanced equation for this reaction. The reactants are HCl and Zn and the products are ZnCl2(aq), and H2(g). For how to write a balanced equation, see the Related Questions to the left. Then, convert the grams of Zn into moles of Zn. To do that, see the Related Questions to the left. Then use stoichiometry to determine how many moles of HCl are necessary to react with that number of moles of Zn. See the Related Questions to the left for how to solve stoichiometry problems. Finally, determine how many milliliters of solution you need to get that many moles of HCl. To do that, use this equation: number of moles = number of liters * molarity
Balanced equation is: Zn(s) + I2(s) --> ZnI2(s) Pick any initial mass of Zn and I2, and convert those masses to moles: 100 g Zn / 63.4 g/mol = 1.58 moles 100 g I2 / 253.8 g/mol = 0.394 mol I2 Since these two react in a 1:1 ratio, you will have used 0.394 moles of Zn to react with the 0.394 moles of I2. That would leave 1.58 - 0.394 = 1.186 moles of Zn unreacted. So, the fraction of the original zinc remaining would be 1.186 / 1.58 = .75 by sonu gupta
Zn + S ==> ZnSmoles Zn = 25 g x 1 mol/65.4 g = 0.38 moles Znmoles S = 300 g x 1 mol/32 g = 9.38 moles Smoles S in excess = 9.38 - 0.38 = 9.00 molesmass S in excess = 9 moles x 32g/mol = 288 g = 280 g (to 2 significant figures)
I assume zinc limits in this reaction. Balanced equation. Zn + 2HCl -> ZnCl2 + H2 184 grams Zn (1 mole Zn/65.41 grams)(1 mole ZnCl2/1 mole Zn) = 2.81 moles zinc chloride produced ==========================
The atomic weight of zinc is 65.4 g/molTo convert grams to moles:moles Zn= 64.0 g Zn1 mol = 0.979 mol Zn65.4 gMultiply by moles per gram. Grams cancel out
Zn + 2HCl ---> H2 + ZnCl2n = 0.112mw = 65.39g = 7.35From these values (they are for the zinc), you can see that only .112 mol of Zn reacted.To find the number of moles of HCl, we use n = cV.n = 1.2 x 0.5= 0.6 x 2 mols= 1.2 mol HCl.From this we can see that zinc is the limiting reagent and that 0.112 mol of H2 gas was produced.g = mw x ng = 2.016 x 0.112= 0.225792 grams.