The balanced equation for the neutralisation reaction is as follows
Ca(OH)₂ + H₂SO₄ ---> CaSO₄ + 2H₂O
stoichiometry of Ca(OH)₂ to H₂SO₄ is 1:1
equivalent number of acid reacts with base
number of H₂SO₄ mol reacting - 2 mol
according to molar ratio of 1:1
number of Ca(OH)₂ mol = number of H₂SO₄ moles
therefore number of Ca(OH)₂ moles required - 2 mol
Answer:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
Explanation:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
The balanced chemical reaction is written as:
<span>4C(s) + S8(s) → 4CS2(l)
We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.
</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8
The limiting reactant would be S8. We use this amount to calculate.
0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
The density is 1.12161 g/ml
Answer:
a) 0,5
Explanation:
If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5
