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From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
34g C * ( 1 mol / 12.0107 ) * ( 1 mol H2 / 1 mol C ) * ( <span>2.01588 g / 1 mol H2 ) = 5.70657164028741 g H2 = 5.7 g H2
Convert grams of C to moles of C using the given amount of grams and the molar mass ( 12.0107 g/mol ).
Gather the mole ratio from the coefficients in the balanced equation and multiply by the ratio.
Convert moles of H2 to grams of H2 </span> using the given amount of grams and the molar mass ( 2.01588 g/mol )<span>.
Revise your answer to have the correct number of significant figures. </span>