Question

for c3h8 + o2 -->co2 + h2o, the coeff for balanced equation are _, and if the reaction is with 14.8 g of c3h8 and 3.44 g o2, the limiting reagent is _. a 1,5,3,4, oxygen b 1,3,2,4, oxygen c 1,5,3,4 c3h8

Explanation:

Step1: Balance the chemical equation

The un - balanced equation is $C_3H_8+O_2
ightarrow CO_2 + H_2O$.
For carbon: There are 3 carbon atoms in $C_3H_8$, so we put a 3 in front of $CO_2$. For hydrogen: There are 8 hydrogen atoms in $C_3H_8$, so we put a 4 in front of $H_2O$. Now the right - hand side has $3\times2 + 4\times1=10$ oxygen atoms. So we put a 5 in front of $O_2$. The balanced equation is $C_3H_8 + 5O_2
ightarrow3CO_2+4H_2O$, and the coefficients are 1, 5, 3, 4.

Step2: Calculate the moles of reactants

The molar mass of $C_3H_8$ is $M_{C_3H_8}=3\times12 + 8\times1=44\ g/mol$. The number of moles of $C_3H_8$, $n_{C_3H_8}=\frac{m_{C_3H_8}}{M_{C_3H_8}}=\frac{14.8\ g}{44\ g/mol}\approx0.336\ mol$.
The molar mass of $O_2$ is $M_{O_2}=32\ g/mol$. The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{3.44\ g}{32\ g/mol}=0.1075\ mol$.
From the balanced equation, the mole ratio of $C_3H_8$ to $O_2$ is 1:5.
For 0.336 mol of $C_3H_8$, the required moles of $O_2$ is $n_{O_2\ required}=0.336\ mol\times5 = 1.68\ mol$. But we only have 0.1075 mol of $O_2$. So $O_2$ is the limiting reagent.

Answer:

A. 1,5,3,4, oxygen