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Could you explain and solve these two limiting reactant problems for me please?
_ indicates a subscript ok? I already balanced them. And how many grams is typed underneath.
1) B_2Br_6 + 6HNO_3 ------> 2B(NO_3)_3 + 6HBr
2.0g 3.0g
2) CH_4 + 2O_2 ------> CO_2 + 2H_2O
3.15g 2.98g
OOPS.. it edited my grams.
B_Br_6 is 2.0g.
6HNO_3 is 3.0g
CH_4 is 3.15g.
2O_2 is 2.98g.
1 Answer
- Roger the MoleLv 78 years agoFavorite Answer
1)
B2Br6 + 6 HNO3 → 2 B(NO3)3 + 6 HBr
(2.0 g B2Br6) / (501.0480 g B2Br6/mol) = 0.003992 mol B2Br6
(3.0 g HNO3) / (63.01296 g HNO3/mol) = 0.04761 mol HNO3
0.003992 mole of B2Br6 would react completely with 0.003992 x (6/1) = 0.023952 mole of HNO3, but there is more HNO3 present than that, so HNO3 is in excess and B2Br6 is the limiting reactant.
You didn't ask what was to be found, so as a guess:
(0.003992 mol B2Br6) x (2 mol B(NO3)3 / 1 mol B2Br6) x (196.8267 g B(NO3)3/mol) =
1.6 g B(NO3)3
2)
CH4 + 2 O2 → CO2 + 2 H2O
(3.15 g CH4) / (16.04257 g CH4/mol) = 0.19635 mol CH4
(2.98 g O2) / (31.99886 g O2/mol) = 0.093128 mol O2
0.093128 mole of O2 would react completely with 0.093128 x (1/2) = 0.046564 mole of CH4, but there is more CH4 present than that, so CH4 is in excess and O2 is the limiting reactant.
Again, guessing about what was to be found:
(0.093128 mol O2) x (1 mol CO2 / 2 mol O2) x (44.00964 g CO2/mol) = 2.05 g CO2
