From the information below, determine the mass of substance \(C\) that will be formed if \(45.0\) grams of substance \(A\) reacts with \(23.0\) grams of substance \(B\). (Assume that the reaction between \(A\) and \(B\) goes to completion.) a. Substance \(A\) is a gray solid that consists of an alkaline earth metal and carbon ( \(37.5 \%\) by mass). It reacts with substance \(B\) to produce substances \(C\) and \(D .\) Forty million trillion formula units of \(A\) have a mass of \(4.26\) milligrams. b. \(47.9\) grams of substance \(B\) contains \(5.36\) grams of hydrogen and \(42.5\) grams of oxygen. c. When \(10.0\) grams of \(C\) is burned in excess oxygen, \(33.8\) grams of carbon dioxide and \(6.92\) grams of water are produced. \(\mathrm{A}\) mass spectrum of substance \(C\) shows a parent molecular ion with a mass-to-charge ratio of 26 . d. Substance \(D\) is the hydroxide of the metal in substance \(A\).

: From the given information: Substance A consists of an alkaline earth metal (M) and carbon (C). Substance B consists of hydrogen (H) and oxygen (O). Substance C forms carbon dioxide (CO_2) and water (H_2O) when burned. Substance D is the hydroxide of the metal in substance A (M(OH)_2).

: a. For substance A: The mass of 40 million trillion (40*10^12) formula units of A is 4.26 mg which is equal to 4.26*10^-3 g. The molar mass of A can be calculated as: \(Molar\:mass\:of\:A=\frac{4.26*10^-3\:g}{40*10^{12}}\) Solving, we get the molar mass of A as 106.5 g/mol. Since substance A contains 37.5% of carbon by mass, \(Mass\:of\:carbon=\frac{37.5}{100}*106.5\:g/mol\ = 39.9375\:g/mol\) The mole ratio of M to C in A is given by : \(\frac{Molar\:mass\:of\:M}{Molar\:mass\:of\:C} = \frac{106.5\:g/mole -39.9375\:g/mole}{12.01\:g/mole}\) Solving, we get the mole ratio as 5.52, which approximates to 6. Thus, the empirical formula of A is MC_6. b. For substance B: 47.9 grams of B contains 5.36 grams of H and 42.5 grams of O. Number of moles of H: \(\frac{5.36\:g}{1.008\:g/mol} = 5.32\:mol\) Number of moles of O: \(\frac{42.5\:g}{16.00\:g/mol} = 2.656\:mol\) \(Mole\:ratio\: of\: H\ to\: O\ in\: B = \frac{5.32\:moles}{2.656\:moles} = 2\) Therefore, the empirical formula of B is H_2O_2. c. For substance C: Since substance C has a parent molecular mass-to-charge ratio of 26, we know the molar mass of C is 26 g/mol. When 10 grams of C is burned, it forms 33.8 grams of CO_2 and 6.92 grams of H_2O. We can deduce the empirical formula of C by calculating the moles of carbon and hydrogen: Moles of carbon in CO_2: \(\frac{33.8\:g}{44.01\:g/mol} = 0.768\:mol\) Moles of hydrogen in H_2O: \(\frac{6.92\:g}{18.015\:g/mol} = 0.384\:mol\) From these values, we see that the mole ratio of carbon to hydrogen in substance C is 2:1. Hence, the empirical formula of C is CH_2.

: Now we have the following equation: \(MC_6 + H_2O_2 \rightarrow CH_2 + M(OH)_2\) To balance the equation: \(3MC_6 + 6H_2O_2 \rightarrow 6CH_2 + 3M(OH)_2\)

: Given, 45.0 grams of A and 23.0 grams of B. Calculate the moles of A and B: Moles of A: \(\frac{45.0\:g}{106.5\:g/mol} = 0.422\:mol\) Moles of B: \(\frac{23.0\:g}{34.014\:g/mol} = 0.676\:mol\) The stoichiometric ratio for A to B from the balanced equation: \(\frac{Moles\:of\:A}{Moles\:of\:B} = \frac{0.422\:mol}{0.676\:mol} = 0.625\) Comparing the stoichiometric ratio to the actual mole ratio from the balanced equation: \(\frac{Moles\:of\:A}{Moles\:of\:B}=\frac{3\:mol}{6\:mol} = 0.5\) Since 0.625 > 0.5, A is in excess, and B is the limiting reactant.

: Since B is the limiting reactant, we can now calculate the mass of substance C produced by the reaction. Using the balanced equation: 6 mol of C produced from 6 mol of B, so moles of C formed = 0.676 mol. Mass of C: \(0.676\:mol * 26\:g/mol = 17.576\:g\) Therefore, the mass of substance C that will be formed is \(17.6\:g\).