A sample contain 33.42g of metal pellets are poured on to a graduated cylinder initially containing 12.7ml of water, causing the
1 answer:
Answer:
3. 76 g/ml^3
Explanation:
We are given the mass of the metal pellets which is 33.4 grams. To calculate the density of the metal, we need to divide the mass by the volume. The volume of the metal is not directly given so we need to find it ourselves. The water was initially 12.7 ml and then it rose to 21.6 ml after placing the metal pellets. This indicate that the volume of the metal pellets is 8.9 ml.
We can now use the density formula.
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This is an incomplete question, here is a complete question.
Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.
Answer : The volume of hydrogen gas that will be collected is 1.85 L
Explanation :
First we have to calculate the number of moles of aluminium.
Given mass of aluminium = 1.35 g
Molar mass of aluminium = 27 g/mol
The given chemical reaction is:
As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.
Thus, aluminium is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
2 moles of aluminium produces 3 moles of hydrogen gas
So, 0.005 moles of aluminium will produce = of hydrogen gas
Now we have to calculate the mass of helium gas by using ideal gas equation.
PV = nRT
where,
P = Pressure of hydrogen gas = 743 Torr
V = Volume of the helium gas = ?
n = number of moles of hydrogen gas = 0.075 mol
R = Gas constant =
T = Temperature of hydrogen gas =
Now put all the given values in above equation, we get:
Hence, the volume of hydrogen gas that will be collected is 1.85 L