Answer:
12 moles of CO₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
CO₂ + H₂O —> H₂CO₃
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Finally, we shall determine the number of moles of CO₂ that will dissolve in water to produce 12 moles of H₂CO₃. This can be obtained as follow:
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Therefore, 12 moles of CO₂ will also dissolve in water to produce 12 moles of H₂CO₃.
Thus, 12 moles of CO₂ is required.
Block and tackle d is the best answer
The molecules of hydrogen gas that are formed is when 48.7 g of sodium are added to water is 6.375 x 10²³ molecules
<u><em>calculation</em></u>
2 Na +2H₂O → 2 NaOH +H₂
Step 1: find the moles of sodium (Na)
moles =mass÷ molar mass
from periodic table the molar mass of Na = 23 g/mol
moles= 48.7 g÷ 23 g/mol =2.117 moles
Step 2:use the mole ratio to determine the moles of H₂
from given equation Na:H₂ is 2:1
therefore the moles of H₂ = 2.117 moles x 1/2=1.059 moles
Step 3: find the molecules of H₂ using the Avogadro's law
According to Avogadro's law 1 mole = 6.02 x 10²³ molecules
1.059 moles = ? molecules
by cross multiplication
= [(1.059 moles x 6.02 x10²³ molecules) / 1 mole] =6.375 x 10²³ molecules
Answer:
83.9g of sulfuric acid is the minimum mass you would need
1.73g of hydrogen would be produced
Explanation:
Based on the reaction:
2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)
2 moles of solid aluminium react with 3 moles of sulfuric acid. Also, two moles of Al produce 3 moles of hydrogen gas.
15.4g of Al are:
15.4g Al × (1mol / 26.98g) = 0.571 moles of Al.
Moles of sulfuric acid:
0.571 moles Al × (3 mol H₂SO₄ / 2 mol Al) = 0.8565 moles H₂SO₄
In grams:
0.8565 moles H₂SO₄ × (98g / 1mol) = <em>83.9g of sulfuric acid is the minimum mass you would need</em>
In the same way, moles of hydrogen produced are:
0.571 moles Al × (3 mol H₂ / 2 mol Al) = 0.8565 moles H₂
In grams:
0.8565 moles H₂ × (2.015g / 1mol) = <em>1.73g of hydrogen would be produced</em>
