the equation represents the combustion of sucrose (c₁₂h₂₂o₁₁). c₁₂h₂₂o₁₁ + 12o₂ → 12co₂ + 11h₂o. if there…

the equation represents the combustion of sucrose (c₁₂h₂₂o₁₁). c₁₂h₂₂o₁₁ + 12o₂ → 12co₂ + 11h₂o. if there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction? 0.029 mol 0.250 mol 0.351 mol 3.00 mol

the equation represents the combustion of sucrose (c₁₂h₂₂o₁₁). c₁₂h₂₂o₁₁ + 12o₂ → 12co₂ + 11h₂o. if there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction? 0.029 mol 0.250 mol 0.351 mol 3.00 mol

Answer

Explanation:

Step1: Calculate molar - mass of sucrose

The molar - mass of $C_{12}H_{22}O_{11}$: The atomic mass of $C = 12.01\ g/mol$, $H = 1.01\ g/mol$, $O = 16.00\ g/mol$. $M=(12\times12.01 + 22\times1.01+11\times16.00)\ g/mol=(144.12 + 22.22+176.00)\ g/mol = 342.34\ g/mol$.

Step2: Calculate moles of sucrose

Use the formula $n=\frac{m}{M}$, where $m = 10.0\ g$ (mass of sucrose) and $M = 342.34\ g/mol$ (molar - mass of sucrose). $n=\frac{10.0\ g}{342.34\ g/mol}\approx0.029\ mol$.

Answer:

0.029 mol