P=0.0902 g/l
v=22.4 l/mol (stp)
M=vp
M=22.4 l/mol * 0.0902 g/l=2.020 g/mol
M=2.020 g/mol
River Banks? I'm not completely sure but hope this helps!
Answer:
5.7 moles of O2
Explanation:
We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:
2KClO3 —> 2KCl + 3O2
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.
Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.
Thus, 5.7 moles of O2 were obtained from the reaction.
<span>Answer:
3 N2H4 (l) = 4 NH3 (g)+ N2 (g)
2 N2H4 (l) + N2O4 (g) =3 N2 (g)+ 4 H2O (g)</span>
