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Tomtit [17]
3 years ago
9

Determine the number of moles in 4.00x10^2 g of cesium iodide

Chemistry
1 answer:
Tema [17]3 years ago
3 0
<span>Given the value for S2- as 3.42 x 10-17 M find the value of Ag+. (2:1 ratio). Ag2S W ... 4.00 x10 M ... First,calculate the number of moles in solution by multiplying volume and .... To find the number of grams of oxygen: 100 – (31.7 + 5.30) = 63.0.</span><span>
</span>
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If you had 6 moles of CaCl2 and 5 moles of Na3PO4, which of these would be the limiting and excess reactant
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Na3PO4 is excess reactant, CaCl2 is limiting reactant.

Explanation:

                                      3CaCl2 + 2Na3PO4 ---> Ca3(PO4)2 + 6NaCl

from reaction :              3 mol        2 mol

given:                            6 mol         5 mol (X)

X = (6*2)/3 = 4 mol Na3PO4

For 6 mol CaCl2 we need 4 mol Na3PO4, but  we have 5 mol Na3PO4,

Na3PO4 is excess reactant, so CaCl2 is limiting reactant.

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