(4.99 g) / (3.65 g/mL) = 1.37 mL
The answer to your question is 1.37ml.
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
The answer is C because you don’t have to worry about the number in front of the decimal unless it is something greater than zero. A- Doesn’t have there sig figs. B- The zero after the 5 doesn’t change it. D- The three at the end doesn’t matter because it does not round it up.
