2C4H10 + 13O2 → 8CO2 + 10H2O Since the equation is balanced, we can set up a proportion: 13 moles of O2 react with 2 moles of C4H10x moles of O2 react with 0.425 moles of C4H10 13 → 2x → 0.425 x = 13 * 0.425 / 2 = 2.7625 <span>2.7625 moles of O2 react with 0.425 moles of C4H10</span>
A) Ca(OH)2 + CO2 —> CaCO3 + H2O
B) when Ca(OH)2 is reacted with CO2, the CaCO3 produced is a precipitate which turns the solution milky
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:

Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g
Order.
Hope this helps! :)
The drawing is correct because there are 12 atoms of each type on each side of the arrow.