0.010 moles of carbon c4h10 reacts with oxygon as in the queation 1.76g of carbon dioxide and 0.90 of water are produced. use this information ti work out the balancing numbers for carbon dioxide and water

The coefficients are 2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

Step 1. Gather all the information in one place.

M_r:       58.12    32.00     44.01    18.02

             aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O

m/g:                                      1.76      0.90

n/mol:  0.010

Step 2. Calculate the mass of C₄H₁₀.

Mass = 0.010 mol C₄H₁₀ × (58.12 g C₄H₁₀/1 mol C₄H₁₀) = 0.581 g C₄H₁₀

Step 3. Calculate the mass of O₂

Mass of C₄H₁₀ + mass of O₂ = mass of CO₂ + mass of H₂O

0.581 g + x g = 1.76 g + 0.90 g

x = 1.76 + 0.90 - 0.581 = 2.079

Our information now has the form:

M_r:       58.12   32.00    44.01    18.02

           aC₄H₁₀ + bO₂ ⟶ cCO₂ + dH₂O

m/g:     0.581    2.079      1.76      0.90

n/mol:  0.010

Step 4. Calculate the moles of each compound.

Moles of O₂ = 2.079 g O₂ × (1 mol O₂/32.00 g O₂) = 0.064 97 mol O₂

Moles of CO₂ = 1.76 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 0.040 00 mol CO₂

Moles of H₂O = 0.90 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.0499 mol H₂O

Our information now has the form:

           aC₄H₁₀ +    bO₂ ⟶   cCO₂ +    dH₂O

n/mol:  0.010   0.064 97  0.040 00  0.0499

Step 5: Calculate the molar ratios of all the compounds.

a:b:c:d = 0.010:0.064 97:0.040 00:0.0499 = 1:6.497:4.000:4.99

= 2 :12.99:8.00:9.98 ≈ 2:13:8:10

∴ a = 2; b = 13; c = 8; d = 10

The balanced equation is

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2