Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.

2C(s) + 3H2(g) = C2H6(g) Explanation: In balancing chemical equations, the reactant must be equal to the product. I.e law of conservation of matter must hold. 2 moles of carbon would react with 3 moles of hydrogen to give 1 mole of methane 2C(s) + 3H2(g) = C2H6(g)

2C4H10(g) + 13O2(g)>8CO2(g) + 10H2O(g) Explanation: From the general formula for combustion of alkanes: CnH2n+2 + 3n+1/2 O2> nCO2 + (n+1)H2O We have: C4H10(g) + 13/2 O2(g) > 4 CO2(g) + 5H2O(g) Now we can multiply by 2 to obtain the lowest whole number coefficients 2C4H10(g) + 13O2 (g) >8CO2(g) + 10 H2O(g) We had to multiply by two because it is the integer that will give us the lowest coefficient. Multiplying through by 2 will give the expected result.

6H2+P4–>4PH3 Explanation:

P₄  +  4NaOH  + 2H₂O  →  2PH₃  +  2Na₂HPO₃   Explanation: The given reaction expression is given as:               P₄  +  NaOH  + H₂O  →  PH₃  +  Na₂HPO₃   To solve this problem, now we put coefficients a, b, c, d and e and we use a mathematical approach to solve this problem;               aP₄  +  bNaOH  + cH₂O  →  dPH₃  +  eNa₂HPO₃   Conserving P: 4a  = d + e                   Na: b  = 2e                   O:  b + c  = 3e                   H:   b + 2c  = 3d + e let b  = 1, e  = , c  =  , d =  , a =   Multiply through by 4; b  = 4, e  = 2, c  = 2, d  = 2, a  = 1         P₄  +  4NaOH  + 2H₂O  →  2PH₃  +  2Na₂HPO₃

1. C(2)+H2(1) -> C2H6(1) 2. NH3(2)+O2(3)-> HCN(2)+H2O(3) I am not sure about the second one.

Answer 6

2C(s) + 3H2(g) –> C2H6(g)

Answer 7

2Fe₂O₃(s) → 4Fe(s) + 3O₂(g) Explanation: The unbalanced reaction is: Fe₂O₃(s) → Fe(s) + O₂(g) To balance the oxygen atoms, we put a 2 before Fe₂O₃ and a 3 before O₂(g) – so as to make the total of O atoms in both sides 6 -: 2Fe₂O₃(s) → Fe(s) + 3O₂(g) There are four Fe atoms on the left side and only one on the right side, so we put a 4 before Fe(s): 2Fe₂O₃(s) → 4Fe(s) + 3O₂(g) Now the equation is balanced: there is an equal number of both Fe and O atoms in both sides of the equation.

Explanation:

0

Put a 3 on the Br2 and a 2 in front of the IBr3. You will then have 6Br on both sides, 2 I’s on both sides.

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