For the following reaction,the mass of water | vedclass

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(C) The balanced chemical equation is:$2C_{57}H_{110}O_{6(s)} + 163O_{2(g)} 	o 114CO_{2(g)} + 110H_2O_{(l)}$First,calculate the molar mass of $C_{57}

Vedclass · Accepted Answer

$495$

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(C) The balanced chemical equation is:$2C_{57}H_{110}O_{6(s)} + 163O_{2(g)} 	o 114CO_{2(g)} + 110H_2O_{(l)}$First,calculate the molar mass of $C_{57}H_{110}O_6$:$M = (57 	imes 12) + (110 	imes 1) + (6 	imes 16) = 684 + 110 + 96 = 890 \ g/mol$.Calculate the moles of $C_{57}H_{110}O_6$:$n = \frac{445 \ g}{890 \ g/mol} = 0.5 \ mol$.From the stoichiometry,$2 \ mol$ of $C_{57}H_{110}O_6$ produces $110 \ mol$ of $H_2O$.Therefore,$0.5 \ mol$ of $C_{57}H_{110}O_6$ produces:$n(H_2O) = \frac{110}{2} 	imes 0.5 = 55 \ mol$.Mass of $H_2O = 55 \ mol 	imes 18 \ g/mol = 990 \ g$.Wait,re-calculating: $55 	imes 18 = 990$. Let's check the provided options. The calculation $\frac{0.5}{2} = \frac{x}{110}$ gives $x = 27.5 \ mol$. $27.5 	imes 18 = 495 \ g$. The stoichiometry ratio is $2:110$,so $1:55$. $0.5 	imes 55 = 27.5 \ mol$. $27.5 	imes 18 = 495 \ g$.

For the following reaction,the mass of water produced from $445 \ g$ of $C_{57}H_{110}O_6$ is .............. $g$.
$2C_{57}H_{110}O_{6(s)} + 163O_{2(g)} \to 114CO_{2(g)} + 110H_2O_{(l)}$

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For the following reaction,the mass of water produced from $445 \ g$ of $C_{57}H_{110}O_6$ is .............. $g$.$2C_{57}H_{110}O_{6(s)} + 163O_{2(g)} \to 114CO_{2(g)} + 110H_2O_{(l)}$