Answer:
<em>Mg = 24.30 g/mol) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Hint: 1 mole of gas at STP occupies 22.4 L</em>
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
I'd love to help, but you forgot to add the question.
Option B. <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO is the correct answer. HOPE IT HELPS</span>
The final destination to where some of the electrons go to at the end of cellular respiration would be D. Oxygen. Assuming that this aerobic cellular respiration, the final electron acceptor is that of oxygen.
