Percent Yields? Mystery Element.?

Question

a volume of 2.27 mL of SiCl (density = 1.438g/ml) reacts with an excess of H2S (g), giving HSSiCl3 according to the reaction:
SiCl4(l) + H2S (g) --> HSSiCl3 (l) + HCl (g).
If the HCl produced reacts with 8.267 x 10^-3 mol of NaOH according to the reaction:
HCl + NaOH --> NaCl + H2O. What is the percent yeid of HSSiCl3?

If that wasn't hard enough...
A common oven cleaner contains a very caustic metal hydroxide, which we represent as MOH. Reaction of .7134 grams of MOH with HNO requries 4.50 grams to reach the end point.
MOH + HNO3 --> MNO3 + H2O.
what is the element M and what is the formula of the hydroxide..

Any help at all would be greatly appreciated. I have no clue how to even attempt. Thank you for your time!

Answer 1

As a "clue", the reaction formulas relate the moles of reactants and products to each other. After that, you can switch over to weights.

In problem 1,each mole of NaOH requires 1 mole of HCL.
But each mole of HCl formed requires 1 mole of Silicon tetrachloride. And with each mole of HCl formed, a mole of the silicon trichloride hyrdosulfide is formed. This gives us the links to do the problem. The amount of SiCl4 available is 3.26 g (2.27 ml x 1.438 g/ml). The mole weight of SiCl4 is about 172, so we have 0.019 moles. If ALL was converted to (HS)-SiCl3, we would have 0.019 moles of HCl which would have reacted with 0.019 moles of NaOH. The percentage yield is based on the moles reacted/moles available for reaction, or .008267/.019, which is about .43. Thus, the yield is about 43%

In problem 2, the moles of MOH will be equal to the moles of HNO3. Once we know that, the weight of MOH will lead us to its molecular weight. if the weight of M is m g,
(m+17)= molecular weight of MOH, from which m can be found. For instance, if m were about 21, the metal would be sodium, which has an atomic weight of 21.

The problem does not seem well-stated, since the reaction of HNO is not represented by the reaction shown. It's not clear which part of the statement is incorrect.