Answer:
mCO2= 49.6932 kgCO2
Explanation:
Hello! Let's solve this!
First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O
We see that each mole of C3H8 (propane) we get 3 moles of CO2
From the propane volume we can obtain the grams of propane used.
molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane
mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2
mCO2= 49.6932 kgCO2
<span>Mr = 13 g / mol
mass = 5 g
Mol = 5/13 mol :)</span>
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