The laboratory procedure that best illustrate the law of conservation is
heating 100 g of CaCo3 to produce 56 g of CaO (answer C)
<u><em>explanation</em></u>
According to the law of mass conservation , the mass of the reactant must be equal to the mass of the product.
According to option c Heating 100 g CaCO3 to produces 56 g CaO ( 40 +16=56)
The remaining mass = 100-56 = 44 which would the mass of CO2 [ 12 + (16 x2)]= 44 since CaCO3 decomposes to produce CaO and CO2
Therefore the mass of reactant= 100g
mass of product = 56 g +44 g =100
Therefore the laboratory procedure for decomposition of CaCO<em>3</em> illustrate the law of mass conservation since the mass of reactant = mass of product.
Answer:
2HNO3 +Na2CO3 → CO2 + H2O + 2NaNO3
Answer:
4.33 L
Explanation:
Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:
Where in this case:
We <u>input the given data</u>:
- 2.16 L * 0.601 mol = V₂ * 0.300 mol
And <u>solve for V₂</u>:
Answer:
sodium hydroxide is the limiting reactant
Explanation:
The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.
2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr
Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.
For nitrogen tribromide
From the balanced reaction equation;
2 moles of nitrogen tribromide yields 1 mole of nitrogen gas
4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas
For sodium hydroxide;
3 moles of sodium hydroxide yields 1 mole of nitrogen gas
5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas
Therefore, sodium hydroxide is the limiting reactant.
