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 equal the product. In other words, the law of conservation of matter must be respected. 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 the 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) We can now multiply by 2 to obtain the smallest integer 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 smallest coefficient. Multiplying by 2 will give the expected result.

6H2+P4–>4PH3 Explanation:

P₄ + 4NaOH + 2H₂O → 2PH₃ + 2Na₂HPO₃ Explanation: The expression for the given reaction is given by: P₄ + NaOH + H₂O → PH3 + Na₂HPO₃ To solve this problem, we now put the coefficients a, b, c, d and and let’s use the 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 i.e. b = 1, e = , c = , d = , a = Multiply 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) 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, put a 2 in front of Fe₂O₃ and a 3 before O₂(g) – so that the total number of O atoms on both sides is 6 – : 2Fe₂O₃(s) → Fe(s) + 3O₂(g) There are four Fe atoms on the side left and only one on the right side, so we put a 4 in front of Fe(s): 2Fe₂O₃(s) → 4Fe(s) + 3O₂(g) Now the equation is balanced: there is an equal number d Fe and O atoms on both sides of the equation.

Explanation:

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

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