Explanation:
hope this helps please like and mark as brainliest
Answer:
first number is the correct answer
Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
<em>Moles C6H12O6:</em>
650g * (1mol/180.16g) = 3.608 moles C6H12O6
<em>Moles O2:</em>
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
<h3>O2 is limiting reactant</h3>
0.01742919 is the answer because i worked it out
Answer:
moles H₂O = 10
Explanation:
The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g
Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g
The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol
The moles of Na₂CO₃ and H₂O in the sample are:
Na₂CO₃ = 1.42 / 106 = 0.01340 moles
H₂O = 2.417 / 18 = 0.1343
Now using rule of three :
1 mole of Na₂CO₃ has x moles of H₂O
0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O
x = 1 * 0.1343 / 0.01340 = 10
