If I start with 5 grams of $C_{3} H_{8}$ $C_3H_8$ , what is the theoretical yield of water in the equation $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ ?

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If I start with 5 grams of $C_{3} H_{8}$ $C_3H_8$ , what is the theoretical yield of water in the equation $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ ?

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Answer 1 · 2016-03-11T10:41:21

$\approx 8.18 g m$ $~~8.18gm$

Explanation:

the molar mass of $C_{3} H_{8} = 3 \times 12 + 8 \times 1 = 44 g m m o l^{=} 1$ $C_3H_8=3xx12+8xx1=44gmmol^=1$
molar mass of water= $18 g m m o l^{- 1}$ $18gmmol^-1$

As per the equation 44 gm $C_{3} H_{8}$ $C_3H_8$ gives $4 \times 18 = 72 g m$ $4xx18=72gm$ water
1gm $C_{3} H_{8}$ $C_3H_8$ gives $4 \times 18 = \frac{72}{44} g m$ $4xx18=72/44gm$ water
5gm $C_{3} H_{8}$ $C_3H_8$ gives $4 \times 18 = \frac{72}{44} \times 5 \approx 8.18 g m$ $4xx18=72/44xx5~~8.18gm$ water

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If I start with 5 grams of $C_{3} H_{8}$ $C_3H_8$ , what is the theoretical yield of water in the equation $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ ?

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If I start with 5 grams of C3H8C_3H_8, what is the theoretical yield of water in the equation C3H8+5O2→3CO2+4H2OC_3H_8 + 5O_2 -> 3CO_2 + 4H_2O?

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If I start with 5 grams of $C_{3} H_{8}$ $C_3H_8$ , what is the theoretical yield of water in the equation $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ ?