53.5 g MgCl2 /95.21 g/mol = 0.5619 Moles MgCl2. Since the ratio is 3MgCl : 2Na3PO4 you multiply 0.5619 x 2/3 (since you're solving for NaPO, the 2 goes on top) which shows we have .3746 Moles of NaPO. Multiply times it's molar mass so 0.3746x 163.94g/mol= 61.4136 grams NaPO. I believe.
Answer:
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Explanation:
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A) CuBr2
b) Al(NO3)3
c) Ca3(PO4)2
d) Fe2S3
e) HgCl2
f) Mg(C2H3O2)2
If 0.400 moles CO and 0.400 moles O2 completely react, 17.604 grams of CO2 would be produced.
First, let us look at the balanced equation of reaction:

According to the equation, the mole ratio of CO and O2 is 2:1. But in reality, the mole ratio supplied is 1:1. Thus, CO is the limiting reactant while O2 is in excess.
Also from the equation, the ratio of CO consumed to that of CO2 produced is 1:1. Thus, 0.400 moles of CO2 would also be produced from 0.400 moles of CO.
Recall that: mole = mass/molar mass
Therefore, the mass in grams of CO2 that would be produced from 0.400 moles can be calculated as:
Mass = mole x molar mass
= 0.400 x 44.01
= 17.604 grams
More on calculating mass from number of moles can be found here: brainly.com/question/12513822