Answer:
11.6 mol O₂
Explanation:
- C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O
In order to solve this problem we need to <u>convert moles of carbon dioxide (CO₂) into moles of oxygen gas (O₂)</u>. To do so we'll use a conversion factor containing the <em>stoichiometric coefficients</em> of the balanced reaction:
- 7.4 mol CO₂ *
= 11.6 mol O₂
Answer:
80.8 g
Explanation:
First, let's write a balanced equation of this reaction
MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O
Now let's convert grams to moles
We gotta find the weight of MgO
24 + 16 = 40 g/mol
12/40 = 0.3 moles of MgO
We can use this to find out how much Magnesium Nitrate will be formed
0.3 x 1 MgO / 1 Mg(NO₃)₂ = 0.3 moles of Magnesium Nitrate formed
Convert moles to grams
Find the weight of Mg(NO₃)₂ but don't forget that 2 subscript acts as a multiplier of whatever is inside that parenthesis.
24 + 14 x 2 + 16 x 3 x 2 = 148 g/mol
148 x 0.3 = 80.8 g
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
2 FeO + 1 C ===》2 Fe + 1 CO2
Li2S + 2 HNO3 --> 2 LiNO3 + H2S
Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S
Li S + H2 N2 O5 -> Li N2 O5 + H2 S
Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2
Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2
