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The number of moles is given by :
n = m / M
n = ( 4.5 g ) / ( 169.9 g / mol ) = 0.02649 moles <---------------
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How many moles of silver are present in 32.46g of AgNO3?
AgNO3 is 169.87 g/mol. Ag is 107.87 g/mol meaning that Ag is 63.5% of AgNO3. 63.5% of 32.46g is 20.61g. 20.61g / 107.87 g/mol = .191 moles.
How many grams of silver can be produced from the reaction of 92.8 g of silver nitrate with 1.34 g of aluminum?
Oxidation-reduction reaction:Ag^+(aq) + Al(s) ===> Ag(s) + Al^3+ or looked at another way... 3AgNO3(aq) + Al(s) ===> Al(NO3)3(aq) + 3Ag(s) moles AgNO3 present = 92.8 g x 1 mole/170 g =0.546 moles moles Al present = 1.34 g x 1 mole/26.9 g = 0.0498 moles Al is limiting based on mole ratio of 3 AgNO3 : 1 Al moles Ag(s) produced = 0.0498 moles Al x 3 moles Ag/mole Al = 0.1494 moles Ag mass of Ag = 0.1494 moles Ag x 108 g/mole = 16.1 g Ag formed
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgClNaCl + AgNO3 --> AgCl + NaNO31 mole of AgCl = 1 mole of NaClThe sample contained 0.01426 moles of NaCl0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl0.8334 / 0.8421 = 98.97% pureI'm pretty sure this is the correct answer.
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
What is the mass of silver in 3.4g AgNo3?
3.4g AgNO3 @ 107.9 g Ag / 169.9 g/mol AgNO3 = 2.159 g Ag youir answer ( 2 sig figs) : 2.2 grams of Ag
How many moles of silver are present in 32.46g of AgNO3?
AgNO3 is 169.87 g/mol. Ag is 107.87 g/mol meaning that Ag is 63.5% of AgNO3. 63.5% of 32.46g is 20.61g. 20.61g / 107.87 g/mol = .191 moles.
How many moles of silver of ions are presented in 32.46 g of AgNO3?
The number of moles is 0,19.
How many moles of AgCl will be produced from 83.0 g of AgNO3 assuming NaCl is available in excess?
X = 0.489 moles of AgCl produced
How much does 13.0 moles of AgNO3 weigh?
13 multiplied by 169.87 (g/mol) = 2208.3g
How many moles of BaCl2 barium chloride is necessary to react with the 7.5 moles AgNo3 silver nitrate in 2AgNO3 BaCl-- 2AgCl Ba(NO3)2?
3,75 moles barium chloride
How many grams of silver can be produced from the reaction of 92.8 g of silver nitrate with 1.34 g of aluminum?
Oxidation-reduction reaction:Ag^+(aq) + Al(s) ===> Ag(s) + Al^3+ or looked at another way... 3AgNO3(aq) + Al(s) ===> Al(NO3)3(aq) + 3Ag(s) moles AgNO3 present = 92.8 g x 1 mole/170 g =0.546 moles moles Al present = 1.34 g x 1 mole/26.9 g = 0.0498 moles Al is limiting based on mole ratio of 3 AgNO3 : 1 Al moles Ag(s) produced = 0.0498 moles Al x 3 moles Ag/mole Al = 0.1494 moles Ag mass of Ag = 0.1494 moles Ag x 108 g/mole = 16.1 g Ag formed
How many molecules are there in 65 g of silver nitrate (AgNO3)?
Well, because you have 65g of AgNO3, you have .3826 moles of silver nitrate. This is found by dividing the number of grams you have by the molar mass of silver nitrate (169.9g/mol). Once you know how many moles there are you can then multiply by Avogodro's number (6.022x1023) to obtain the number of molecules. In this case it is 2.304x1023 molecules.
What is the mass of 0.118 moles of AgNO3?
Mole / Molar mass = mass = 0.118 [mol] /169.87 [g mol−1] = 20.05 grams
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgClNaCl + AgNO3 --> AgCl + NaNO31 mole of AgCl = 1 mole of NaClThe sample contained 0.01426 moles of NaCl0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl0.8334 / 0.8421 = 98.97% pureI'm pretty sure this is the correct answer.
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
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 mass of silver in 3.4g AgNo3?
3.4g AgNO3 @ 107.9 g Ag / 169.9 g/mol AgNO3 = 2.159 g Ag youir answer ( 2 sig figs) : 2.2 grams of Ag
