Answer:
I remember doing this in 7th,
1. D
2. B or D, more leaning on B though
3. A
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
It mean it consisted of 1 g of lead and 0.077 g of O2.
divide these numbers by molar mas.
1/82=0.012 Pb /0.004 = 3
0.077/16= 0.004 O /0.004 =1
Pb3O
Answer:
There is a production of 11.6 moles of CO₂
Explanation:
The reaction is this:
2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)
2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.
If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)
2 moles of ethane produce 4 moles of dioxide
5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles
Pentane is the correct answer the prefix pent means 5 so 5 hydrogens
