Answer:
heat is the movement from areas of <u>high </u>temperature to areas of <u>low </u>temperature
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
Fe2+:ferrous ion; fe3+ ferric ion are the one which are correctly named(Answer A) Sn2+ are called stannous ion while Sn4+ are known as stannic ions. Pb2+ are called lead (II) ions while Pb4+ are known as lead (IV) ion Co2+ are known was cobalt (II) ions while Co3+ are known as cobalt(IIi) ion.
<h2>
Answer: 131.9 g</h2>
<h3>
Explanation:</h3>
<u>Write a Balanced Equation for the decomposition</u>
CaCO₃ → CaO + CO₂
<u></u>
<u>Find Moles of CO₂ Produced</u>
Since the mole ratio of CaCO₃ to CO₂ is 1 to 1,
the moles of CaCO₃ = moles of CO₂
moles of CaCO₃ = mass ÷ molar mass
= 300 g ÷ 100.087 g/mol
= 2.997 moles
∴ moles of CO₂ = 2.997 moles
<u>Determine Mass of CO₂</u>
Mass = moles × molar mass
= 2.997 mol × 44.01 g/mol
= 131.9 g
<u></u>
<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
