CH3CH2CHO + 2Ag(NH3)2OH = CH3CH2COONH4+2Ag+H20+3NH3
yield of Ag obtained = 75%
calculate minimum mass of propanal that must have been used to form 5g of silver
looks like each mole of propanol produces 2 moles of Ag.
If the reaction yields 75%, then each mole of propanol produces 1.5 moles Ag.
So, how many moles of Ag in 5g?
3/2 that of propanol, convert to g.
still confused
1 mole of Ag = 108 g
So, 5g = 5/108 = 0.0463 moles Ag
Each mole of propanol produces 1.5 moles Ag, so each mole of Ag requires 2/3 moles of propanol (sorry for the 3/2 typo above).
2/3 * .0463 = 0.0309 moles of propanol required.
mol wt of propanol is 58g, so you need 1.79 g
thankyou i appreciate it
thankyou i appreciate it
could u however plz explain how u knew 1 mol of ag =108g
never mind i know now thankyou
To calculate the minimum mass of propanal (CH3CH2CHO) required to produce 5g of silver (Ag), we need to use the stoichiometry of the balanced equation and consider the yield.
First, let's examine the balanced equation:
CH3CH2CHO + 2Ag(NH3)2OH = CH3CH2COONH4 + 2Ag + H2O + 3NH3
From the equation, we can see that for every 2 moles of CH3CH2CHO, we obtain 2 moles of Ag. Therefore, the stoichiometry is 1:1 between the propanal and silver.
Given that the yield of silver obtained is 75%, we can calculate the actual amount of silver produced using the formula:
Actual yield = Theoretical yield × Percent yield/100
Since the percent yield is 75%, the actual yield will be:
Actual yield = 5g × 75/100 = 3.75g
Since the stoichiometry between CH3CH2CHO and Ag is 1:1, the mass of propanal (CH3CH2CHO) used will also be 3.75g.
Therefore, the minimum mass of propanal that must have been used to form 5g of silver is 3.75g.