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The atomic weight of zinc is 65.4 g/mol
To convert grams to moles:
moles Zn= 64.0 g Zn1 mol = 0.979 mol Zn65.4 g
Multiply by moles per gram. Grams cancel out
Wiki User
∙ 13y ago
Wiki User
∙ 13y agothe mole present in 24.4 g OF ZnSO4 is 0.151
Wiki User
∙ 12y ago48. 7 grams ZnS ( 1 mole ZnS/97.48 grams)
= 0.499 moles zinc sulfide
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Wiki User
∙ 9y agoFormula mass of ZnSO4 = 65.4 + 32.1 + 4(16.0) = 161.5
Amount of ZnSO4 in 240g sample = 240/161.5 = 1.49mol
Wiki User
∙ 8y ago3,6 g zinc is equivalent to 0,055 moles.
Wiki User
∙ 8y ago0,.555 kg of ZnSO4 (anhydrous) is equal to 3,437 moles.
Wiki User
∙ 7y ago286 g of Zn3(PO4)2 is equal to 0,74 moles.
Wiki User
∙ 11y ago2.5
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How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
How many moles are in 125g of Zn?
The equivalent of 125 g zinc is 1, 91 moles.
How many mole in 3.65 g Zn?
3.65(g) / 65.38(g/mol) = 0.0558 (mol)
What is the amount in moles of 3.45 g Zn?
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
How many moles are in 22.5 grams of zinc?
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
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
How many moles are in 125g of Zn?
The equivalent of 125 g zinc is 1, 91 moles.
What is the mass in grams of 0.040 moles Zn?
0,040 moles Zn equal 2,614 g.
How many grams of Zn are in 0.356 moles of zinc?
0,356 moles of zinc contain 23,27 g.
How many mole in 3.65 g Zn?
3.65(g) / 65.38(g/mol) = 0.0558 (mol)
What is the amount in moles of 3.45 g Zn?
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
How many moles are in 22.5 grams of zinc?
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
How many milliliters of 0.248M HCl are needed to react with 1.36 grams of zinc to produce hydrogen?
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
Equal weight of Zinc metal and Iodine are mixed together and the Iodine is completely converted to zinc iodide What fraction of weight of the original Zinc remains unreacted?
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
If you have 25g of Zinc and 300g of Sulfur what is the mass of the excess reactant?
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)
How many moles of hydrogen are produced from the reaction of three moles of zinc with an excess of hydrochloric acid?
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 ==========================
What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCl?
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.
