Mass = moles x molar mass
so mass of 6 moles of h2 is: 6×1×2 = 12g
Percent composition by mass of oxygen =
((16.0*6)/(40.1+2*(14.0+16.0*3)))*100%
= 58.5%
therefore, the answer is D
Answer:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
Explanation:
<u>first step </u>
NO2(g) ------------------------------------> NO(g) + O(g)
<u>second step</u>
NO2(g) + O(g) -----------------------------> NO(g) + O2(g)
The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10