How much water must be added to 36.0 g of srcl2 to produce a solution that is 35.0 wt% srcl2? how much water must be added to 36
1 answer:
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:
It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:
Thus, the mass in excess is:
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<u>Answer:</u> The element represented by M is Strontium.
<u>Explanation:</u>
Let us consider the molar mass of metal be 'x'.
The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol
It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:
The metal atom having molar mass as 87.62/mol is Strontium.
Hence, the element represented by M is Strontium.