The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required
Answers:
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According the VSEPR theory the molecular geometry for CH3+ is triagonal planar
Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.
the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
