Question

question 17
using the equation 2c6h11 + 15o2 → 12co2+ 22h2o, blank 1 grams of co2 would be produced from 5.6 moles of c6h11.
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question 18
using the equation 2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o, blank 1 grams of water would be produced from 3.57 grams of magnesium hydroxide.
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Explanation:

Question 17

Step1: Determine mole - ratio

From the equation $2C_{11}H_{24}+15O_{2}
ightarrow12CO_{2}+22H_{2}O$, the mole - ratio of $C_{11}H_{24}$ to $CO_{2}$ is $2:12 = 1:6$.

Step2: Calculate moles of $CO_{2}$

If we have 5.6 moles of $C_{11}H_{24}$, then the moles of $CO_{2}$ produced is $n_{CO_{2}}=5.6\times6 = 33.6$ moles.

Step3: Calculate mass of $CO_{2}$

The molar mass of $CO_{2}$ is $M = 12+2\times16=44$ g/mol. So the mass of $CO_{2}$ is $m = n\times M=33.6\times44 = 1478.4$ g.

Step1: Calculate moles of $Mg(OH)_{2}$

The molar mass of $Mg(OH)_{2}$ is $M_{Mg(OH)_{2}}=24+(16 + 1)\times2=58$ g/mol. The moles of $Mg(OH)_{2}$ in 3.57 g is $n_{Mg(OH)_{2}}=\frac{3.57}{58}=0.06155$ moles.

Step2: Determine mole - ratio

From the equation $2H_{3}PO_{4}+3Mg(OH)_{2}
ightarrow Mg_{3}(PO_{4})_{2}+6H_{2}O$, the mole - ratio of $Mg(OH)_{2}$ to $H_{2}O$ is $3:6 = 1:2$.

Step3: Calculate moles of $H_{2}O$

The moles of $H_{2}O$ produced is $n_{H_{2}O}=0.06155\times2 = 0.1231$ moles.

Step4: Calculate mass of $H_{2}O$

The molar mass of $H_{2}O$ is $M_{H_{2}O}=2\times1 + 16=18$ g/mol. So the mass of $H_{2}O$ is $m = n\times M=0.1231\times18 = 2.2158$ g.

Answer:

1478.4

Question 18