The amount of W(OH)2 needed would be 448.126 g
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
W(OH)2 + 2 HCl → WCl2 + 2 H2O
The mole ratio of W(OH)2 to HCl is 1:2
Mole of 150g HCl = 150/36.461
= 4.11 moles
Equivalent mole of W(OH)2 = 4.11/2
= 2.06 moles
Mass of 2.06 moles W(OH)2 = 2.06 x 217.855
= 448.188g
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Place a burning splint near the opening of a test tube. If a popping noise occurs, it's probably hydrogen. Place a glowing splint in the test tube, and if it reignites, it could be oxygen. Place a burning splint into a test tube, and if it goes out, it could be carbon dioxide.
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:
12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed
Hope this answers the question.</span>
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
