4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
<span>The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions.
</span><span>Example:
NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle). </span>
The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.
